Suppose $A$ has size $n$ and $k$ distinct eigenvalues, $\scalarlist{\lambda}{k}$. Let $S_i=\set{\vect{x}_{i1},\,\vect{x}_{i2},\,\vect{x}_{i3},\,\ldots,\,\vect{x}_{i\geomult{A}{\lambda_i}}}$, denote a basis for the eigenspace of $\lambda_i$, $\eigenspace{A}{\lambda_i}$, for $1\leq i\leq k$. Then \begin{equation*} S=S_1\cup S_2\cup S_3\cup\cdots\cup S_k \end{equation*} is a set of eigenvectors for $A$. A vector cannot be an eigenvector for two different eigenvalues (see Exercise EE.T20) so $S_i\cap S_j=\emptyset$ whenever $i\neq j$. In other words, $S$ is a disjoint union of $S_i$, $1\leq i\leq k$.
(⇐) The size of $S$ is \begin{align*} \card{S} &=\sum_{i=1}^k\geomult{A}{\lambda_i}&&\text{$S$ disjoint union of $S_i$}\\ &=\sum_{i=1}^k\algmult{A}{\lambda_i}&&\text{Hypothesis}\\ &=n&&\knowl{./knowls/theorem.NEM.knowl}{\text{Theorem NEM}} \end{align*}
We next show that $S$ is a linearly independent set. So we will begin with a relation of linear dependence on $S$, using doubly-subscripted scalars and eigenvectors, \begin{align*} \zerovector= &\left(a_{11}\vect{x}_{11}+a_{12}\vect{x}_{12}+\cdots+a_{1\geomult{A}{\lambda_1}}\vect{x}_{1\geomult{A}{\lambda_1}}\right)+\\ &\left(a_{21}\vect{x}_{21}+a_{22}\vect{x}_{22}+\cdots+a_{2\geomult{A}{\lambda_2}}\vect{x}_{2\geomult{A}{\lambda_2}}\right)+\\ &\left(a_{31}\vect{x}_{31}+a_{32}\vect{x}_{32}+\cdots+a_{3\geomult{A}{\lambda_3}}\vect{x}_{3\geomult{A}{\lambda_3}}\right)+\\ &\quad\quad\vdots\\ &\left(a_{k1}\vect{x}_{k1}+a_{k2}\vect{x}_{k2}+\cdots+a_{k\geomult{A}{\lambda_k}}\vect{x}_{k\geomult{A}{\lambda_k}}\right) \end{align*}
Define the vectors $\vect{y}_i$, $1\leq i\leq k$ by \begin{align*} \vect{y}_1&=\left(a_{11}\vect{x}_{11}+a_{12}\vect{x}_{12}+a_{13}\vect{x}_{13}+\cdots+a_{\geomult{A}{1\lambda_1}}\vect{x}_{1\geomult{A}{\lambda_1}}\right)\\ \vect{y}_2&=\left(a_{21}\vect{x}_{21}+a_{22}\vect{x}_{22}+a_{23}\vect{x}_{23}+\cdots+a_{\geomult{A}{2\lambda_2}}\vect{x}_{2\geomult{A}{\lambda_2}}\right)\\ \vect{y}_3&=\left(a_{31}\vect{x}_{31}+a_{32}\vect{x}_{32}+a_{33}\vect{x}_{33}+\cdots+a_{\geomult{A}{3\lambda_3}}\vect{x}_{3\geomult{A}{\lambda_3}}\right)\\ &\quad\quad\vdots\\ \vect{y}_k&=\left(a_{k1}\vect{x}_{k1}+a_{k2}\vect{x}_{k2}+a_{k3}\vect{x}_{k3}+\cdots+a_{\geomult{A}{k\lambda_k}}\vect{x}_{k\geomult{A}{\lambda_k}}\right) \end{align*}
Then the relation of linear dependence becomes \begin{align*} \zerovector&=\vect{y}_1+\vect{y}_2+\vect{y}_3+\cdots+\vect{y}_k \end{align*}
Since the eigenspace $\eigenspace{A}{\lambda_i}$ is closed under vector addition and scalar multiplication, $\vect{y}_i\in\eigenspace{A}{\lambda_i}$, $1\leq i\leq k$. Thus, for each $i$, the vector $\vect{y}_i$ is an eigenvector of $A$ for $\lambda_i$, or is the zero vector. Recall that sets of eigenvectors whose eigenvalues are distinct form a linearly independent set by Theorem EDELI. Should any (or some) $\vect{y}_i$ be nonzero, the previous equation would provide a nontrivial relation of linear dependence on a set of eigenvectors with distinct eigenvalues, contradicting Theorem EDELI. Thus $\vect{y}_i=\zerovector$, $1\leq i\leq k$.
Each of the $k$ equations, $\vect{y}_i=\zerovector$, is a relation of linear dependence on the corresponding set $S_i$, a set of basis vectors for the eigenspace $\eigenspace{A}{\lambda_i}$, which is therefore linearly independent. From these relations of linear dependence on linearly independent sets we conclude that the scalars are all zero, more precisely, $a_{ij}=0$, $1\leq j\leq\geomult{A}{\lambda_i}$ for $1\leq i\leq k$. This establishes that our original relation of linear dependence on $S$ has only the trivial relation of linear dependence, and hence $S$ is a linearly independent set.
We have determined that $S$ is a set of $n$ linearly independent eigenvectors for $A$, and so by Theorem DC is diagonalizable.
(⇒) Now we assume that $A$ is diagonalizable. Aiming for a contradiction (Proof Technique CD), suppose that there is at least one eigenvalue, say $\lambda_t$, such that $\geomult{A}{\lambda_t}\neq\algmult{A}{\lambda_t}$. By Theorem ME we must have $\geomult{A}{\lambda_t}<\algmult{A}{\lambda_t}$, and $\geomult{A}{\lambda_i}\leq\algmult{A}{\lambda_i}$ for $1\leq i\leq k$, $i\neq t$.
Since $A$ is diagonalizable, Theorem DC guarantees a set of $n$ linearly independent vectors, all of which are eigenvectors of $A$. Let $n_i$ denote the number of eigenvectors in $S$ that are eigenvectors for $\lambda_i$, and recall that a vector cannot be an eigenvector for two different eigenvalues (Exercise EE.T20). $S$ is a linearly independent set, so the subset $S_i$ containing the $n_i$ eigenvectors for $\lambda_i$ must also be linearly independent. Because the eigenspace $\eigenspace{A}{\lambda_i}$ has dimension $\geomult{A}{\lambda_i}$ and $S_i$ is a linearly independent subset in $\eigenspace{A}{\lambda_i}$, Theorem G tells us that $n_i\leq\geomult{A}{\lambda_i}$, for $1\leq i\leq k$.
Putting all these facts together gives, \begin{align*} n &=n_1+n_2+n_3+\cdots+n_t+\cdots+n_k &&\knowl{./knowls/definition.SU.knowl}{\text{Definition SU}}\\ &\leq\geomult{A}{\lambda_1}+\geomult{A}{\lambda_2}+\geomult{A}{\lambda_3}+\cdots+\geomult{A}{\lambda_t}+\cdots+\geomult{A}{\lambda_k} &&\knowl{./knowls/theorem.G.knowl}{\text{Theorem G}}\\ &<\algmult{A}{\lambda_1}+\algmult{A}{\lambda_2}+\algmult{A}{\lambda_3}+\cdots+\algmult{A}{\lambda_t}+\cdots+\algmult{A}{\lambda_k}&&\knowl{./knowls/theorem.ME.knowl}{\text{Theorem ME}}\\ &=n &&\knowl{./knowls/theorem.NEM.knowl}{\text{Theorem NEM}} \end{align*}
This is a contradiction (we cannot have $n<n$!) and so our assumption that some eigenspace had less than full dimension was false.