We begin with a proof by induction (Proof Technique I) of the first statement in the conclusion of the theorem. We use induction on the dimension of $V$ to show that if $\ltdefn{T}{V}{V}$ is a linear transformation, then there is a basis $B$ for $V$ such that the matrix representation of $T$ relative to $B$, $\matrixrep{T}{B}{B}$, is an upper triangular matrix.
To start suppose that $\dimension{V}=1$. Choose any nonzero vector $\vect{v}\in V$ and realize that $V=\spn{\set{\vect{v}}}$. Then $\lt{T}{\vect{v}}=\beta\vect{v}$ for some $\beta\in\complexes$, which determines $T$ uniquely (Theorem LTDB). This description of $T$ also gives us a matrix representation relative to the basis $B=\set{\vect{v}}$ as the $1\times 1$ matrix with lone entry equal to $\beta$. And this matrix representation is upper triangular (Definition UTM).
For the induction step let $\dimension{V}=m$, and assume the theorem is true for every linear transformation defined on a vector space of dimension less than $m$. By Theorem EMHE (suitably converted to the setting of a linear transformation), $T$ has at least one eigenvalue, and we denote this eigenvalue as $\lambda$. (We will remark later about how critical this step is.) We now consider properties of the linear transformation $\ltdefn{T-\lambda I_V}{V}{V}$.
Let $\vect{x}$ be an eigenvector of $T$ for $\lambda$. By definition $\vect{x}\neq\zerovector$. Then \begin{align*} \lt{\left(T-\lambda I_V\right)}{\vect{x}} &=\lt{T}{\vect{x}}-\lambda\lt{I_V}{\vect{x}} &&\knowl{./knowls/theorem.VSLT.knowl}{\text{Theorem VSLT}}\\ &=\lt{T}{\vect{x}}-\lambda\vect{x} &&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\lambda\vect{x}-\lambda\vect{x} &&\knowl{./knowls/definition.EELT.knowl}{\text{Definition EELT}}\\ &=\zerovector &&\knowl{./knowls/property.AI.knowl}{\text{Property AI}} \end{align*}
So $T-\lambda I_V$ is not injective, as it has a nontrivial kernel (Theorem KILT). With an application of Theorem RPNDD we bound the rank of $T-\lambda I_V$, \begin{align*} \rank{T-\lambda I_V}&=\dimension{V}-\nullity{T-\lambda I_V}\leq m-1 \end{align*}
Let $W$ be the subspace of $V$ that is the range of $T-\lambda I_V$, $W=\rng{T-\lambda I_V}$, and define $k=\dimension{W}\leq m-1$. We define a new linear transformation $S$, on $W$, \begin{align*} \ltdefn{S}{W}{W},\quad\lt{S}{\vect{w}}=\lt{T}{\vect{w}} \end{align*}
This does not look we have accomplished much, since the action of $S$ is identical to the action of $T$. For our purposes this will be a good thing. What is different is the domain and codomain. $S$ is defined on $W$, a vector space with dimension less than $m$, and so is susceptible to our induction hypothesis. Verifying that $S$ is really a linear transformation is almost entirely routine, with one exception. Employing $T$ in our definition of $S$ raises the possibility that the outputs of $S$ will not be contained within $W$ (but instead will lie inside $V$, but outside $W$). To examine this possibility, suppose that $\vect{w}\in W$. \begin{align*} \lt{S}{\vect{w}} &=\lt{T}{\vect{w}}\\ &=\lt{T}{\vect{w}}+\zerovector&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\ &=\lt{T}{\vect{w}}+\left(\lambda\lt{I_V}{\vect{w}}-\lambda\lt{I_V}{\vect{w}}\right)&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}}\\ &=\left(\lt{T}{\vect{w}}-\lambda\lt{I_V}{\vect{w}}\right)+\lambda\lt{I_V}{\vect{w}}&&\knowl{./knowls/property.AA.knowl}{\text{Property AA}}\\ &=\left(\lt{T}{\vect{w}}-\lambda\lt{I_V}{\vect{w}}\right)+\lambda\vect{w}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\lt{\left(T-\lambda I_V\right)}{\vect{w}}+\lambda\vect{w} &&\knowl{./knowls/theorem.VSLT.knowl}{\text{Theorem VSLT}} \end{align*}
Since $W$ is the range of $T-\lambda I_V$, $\lt{\left(T-\lambda I_V\right)}{\vect{w}}\in W$. And by Property SC, $\lambda\vect{w}\in W$. Finally, applying Property AC we see by closure that the sum is in $W$ and so we conclude that $\lt{S}{\vect{w}}\in W$. This argument convinces us that it is legitimate to define $S$ as we did with $W$ as the codomain.
$S$ is a linear transformation defined on a vector space with dimension $k$, less than $m$, so we can apply the induction hypothesis and conclude that $W$ has a basis, $C=\set{\vectorlist{w}{k}}$, such that the matrix representation of $S$ relative to $C$ is an upper triangular matrix.
Beginning with the linearly independent set $C$, repeatedly apply Theorem ELIS to add vectors to $C$, maintaining a linearly independent set and spanning ever larger subspaces of $V$. This process will end with the addition of $m-k$ vectors, which together with $C$ will span all of $V$. Denote these vectors as $D=\set{\vectorlist{u}{{m-k}}}$. Then $B=C\cup D$ is a basis for $V$, and is the basis we desire for the conclusion of the theorem. So we now consider the matrix representation of $T$ relative to $B$.
Since the definition of $T$ and $S$ agree on $W$, the first $k$ columns of $\matrixrep{T}{B}{B}$ will have the upper triangular matrix representation of $S$ in the first $k$ rows. The remaining $m-k$ rows of these first $k$ columns will be all zeros since the outputs of $T$ for basis vectors from $C$ are all contained in $W$ and hence are linear combinations of the basis vectors in $C$. The situation for $T$ on the basis vectors in $D$ is not quite as pretty, but it is close.
For $1\leq i\leq m-k$, consider \begin{align*} \vectrep{B}{\lt{T}{\vect{u}_i}} &=\vectrep{B}{\lt{T}{\vect{u}_i}+\zerovector}&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\ &=\vectrep{B}{\lt{T}{\vect{u}_i}+\left(\lambda\lt{I_V}{\vect{u}_i}-\lambda\lt{I_V}{\vect{u}_i}\right)}&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}}\\ &=\vectrep{B}{\left(\lt{T}{\vect{u}_i}-\lambda\lt{I_V}{\vect{u}_i}\right)+\lambda\lt{I_V}{\vect{u}_i}}&&\knowl{./knowls/property.AA.knowl}{\text{Property AA}}\\ &=\vectrep{B}{\left(\lt{T}{\vect{u}_i}-\lambda\lt{I_V}{\vect{u}_i}\right)+\lambda\vect{u}_i}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\vectrep{B}{\lt{\left(T-\lambda I_V\right)}{\vect{u}_i}+\lambda\vect{u}_i} &&\knowl{./knowls/theorem.VSLT.knowl}{\text{Theorem VSLT}}\\ &=\vectrep{B}{\lincombo{a}{w}{k}+\lambda\vect{u}_i} &&\knowl{./knowls/definition.RLT.knowl}{\text{Definition RLT}}\\ &=\colvector{a_1\\a_2\\\vdots\\a_k\\0\\\vdots\\0\\\lambda\\0\\\vdots\\0} &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}} \end{align*}
In the penultimate equality, we have rewritten an element of the range of $T-\lambda I_V$ as a linear combination of the basis vectors, $C$, for the range of $T-\lambda I_V$, $W$, using the scalars $\scalarlist{a}{k}$. If we incorporate these $m-k$ column vectors into the matrix representation $\matrixrep{T}{B}{B}$ we find $m-k$ occurrences of $\lambda$ on the diagonal, and any nonzero entries lying only in the first $k$ rows. Together with the $k\times k$ upper triangular representation in the upper left-hand corner, the entire matrix representation for $T$ is clearly upper triangular. This completes the induction step. So for any linear transformation there is a basis that creates an upper triangular matrix representation.
We have one more statement in the conclusion of the theorem to verify. The eigenvalues of $T$, and their multiplicities, can be computed with the techniques of Chapter E relative to any matrix representation (Theorem EER). We take this approach with our upper triangular matrix representation $\matrixrep{T}{B}{B}$. Let $d_i$ be the diagonal entry of $\matrixrep{T}{B}{B}$ in row $i$ and column $i$. Then the characteristic polynomial, computed as a determinant (Definition CP) with repeated expansions about the first column, is \begin{align*} \charpoly{\matrixrep{T}{B}{B}}{x}&= \left(d_1-x\right) \left(d_2-x\right) \left(d_3-x\right) \cdots \left(d_m-x\right) \end{align*}
The roots of the polynomial equation $\charpoly{\matrixrep{T}{B}{B}}{x}=0$ are the eigenvalues of the linear transformation (Theorem EMRCP). So each diagonal entry is an eigenvalue, and is repeated on the diagonal exactly $\algmult{T}{\lambda}$ times (Definition AME).