We will prove the result by using induction on $p$ (Proof Technique I). To begin, we prove that $T$ has the desired properties when $p=1$. In this case $\vect{u}_1=\vect{v}_1$ and $T=\set{\vect{u}_1}=\set{\vect{v}_1}=S$. Because $S$ and $T$ are equal, $\spn{S}=\spn{T}$. Equally trivial, $T$ is an orthogonal set. If $\vect{u}_1=\zerovector$, then $S$ would be a linearly dependent set, a contradiction.
Suppose that the theorem is true for any set of $p-1$ linearly independent vectors. Let $S=\set{\vectorlist{v}{p}}$ be a linearly independent set of $p$ vectors. Then $S^\prime=\set{\vectorlist{v}{p-1}}$ is also linearly independent. So we can apply the theorem to $S^\prime$ and construct the vectors $T^\prime=\set{\vectorlist{u}{p-1}}$. $T^\prime$ is therefore an orthogonal set of nonzero vectors and $\spn{S^\prime}=\spn{T^\prime}$. Define \begin{equation*} \vect{u}_p=\vect{v}_p -\frac{\innerproduct{\vect{u}_1}{\vect{v}_p}}{\innerproduct{\vect{u}_1}{\vect{u}_1}}\vect{u}_1 -\frac{\innerproduct{\vect{u}_2}{\vect{v}_p}}{\innerproduct{\vect{u}_2}{\vect{u}_2}}\vect{u}_2 -\frac{\innerproduct{\vect{u}_3}{\vect{v}_p}}{\innerproduct{\vect{u}_3}{\vect{u}_3}}\vect{u}_3 -\cdots -\frac{\innerproduct{\vect{u}_{p-1}}{\vect{v}_p}}{\innerproduct{\vect{u}_{p-1}}{\vect{u}_{p-1}}}\vect{u}_{p-1} \end{equation*} and let $T=T^\prime\cup\set{\vect{u}_p}$. We need to now show that $T$ has several properties by building on what we know about $T^\prime$. But first notice that the above equation has no problems with the denominators ($\innerproduct{\vect{u}_i}{\vect{u}_i}$) being zero, since the $\vect{u}_i$ are from $T^\prime$, which is composed of nonzero vectors.
We show that $\spn{T}=\spn{S}$, by first establishing that $\spn{T}\subseteq\spn{S}$. Suppose $\vect{x}\in\spn{T}$, so \begin{equation*} \vect{x}=\lincombo{a}{u}{p} \end{equation*} The term $a_p\vect{u}_p$ is a linear combination of vectors from $T^\prime$ and the vector $\vect{v}_p$, while the remaining terms are a linear combination of vectors from $T^\prime$. Since $\spn{T^\prime}=\spn{S^\prime}$, any term that is a multiple of a vector from $T^\prime$ can be rewritten as a linear combination of vectors from $S^\prime$. The remaining term $a_p\vect{v}_p$ is a multiple of a vector in $S$. So we see that $\vect{x}$ can be rewritten as a linear combination of vectors from $S$, i.e. $\vect{x}\in\spn{S}$.
To show that $\spn{S}\subseteq\spn{T}$, begin with $\vect{y}\in\spn{S}$, so \begin{equation*} \vect{y}=\lincombo{a}{v}{p} \end{equation*}
Rearrange our defining equation for $\vect{u}_p$ by solving for $\vect{v}_p$. Then the term $a_p\vect{v}_p$ is a multiple of a linear combination of elements of $T$. The remaining terms are a linear combination of $\vectorlist{v}{p-1}$, hence an element of $\spn{S^\prime}=\spn{T^\prime}$. Thus these remaining terms can be written as a linear combination of the vectors in $T^\prime$. So $\vect{y}$ is a linear combination of vectors from $T$, i.e. $\vect{y}\in\spn{T}$.
The elements of $T^\prime$ are nonzero, but what about $\vect{u}_p$? Suppose to the contrary that $\vect{u}_p=\zerovector$, \begin{align*} \zerovector&=\vect{u}_p=\vect{v}_p -\frac{\innerproduct{\vect{u}_1}{\vect{v}_p}}{\innerproduct{\vect{u}_1}{\vect{u}_1}}\vect{u}_1 -\frac{\innerproduct{\vect{u}_2}{\vect{v}_p}}{\innerproduct{\vect{u}_2}{\vect{u}_2}}\vect{u}_2 -\frac{\innerproduct{\vect{u}_3}{\vect{v}_p}}{\innerproduct{\vect{u}_3}{\vect{u}_3}}\vect{u}_3 -\cdots -\frac{\innerproduct{\vect{u}_{p-1}}{\vect{v}_p}}{\innerproduct{\vect{u}_{p-1}}{\vect{u}_{p-1}}}\vect{u}_{p-1}\\ &\vect{v}_p= \frac{\innerproduct{\vect{u}_1}{\vect{v}_p}}{\innerproduct{\vect{u}_1}{\vect{u}_1}}\vect{u}_1 +\frac{\innerproduct{\vect{u}_2}{\vect{v}_p}}{\innerproduct{\vect{u}_2}{\vect{u}_2}}\vect{u}_2 +\frac{\innerproduct{\vect{u}_3}{\vect{v}_p}}{\innerproduct{\vect{u}_3}{\vect{u}_3}}\vect{u}_3 +\cdots +\frac{\innerproduct{\vect{u}_{p-1}}{\vect{v}_p}}{\innerproduct{\vect{u}_{p-1}}{\vect{u}_{p-1}}}\vect{u}_{p-1} \end{align*}
Since $\spn{S^\prime}=\spn{T^\prime}$ we can write the vectors $\vectorlist{u}{p-1}$ on the right side of this equation in terms of the vectors $\vectorlist{v}{p-1}$ and we then have the vector $\vect{v}_p$ expressed as a linear combination of the other $p-1$ vectors in $S$, implying that $S$ is a linearly dependent set (Theorem DLDS), contrary to our lone hypothesis about $S$.
Finally, it is a simple matter to establish that $T$ is an orthogonal set, though it will not appear so simple looking. Think about your objects as you work through the following — what is a vector and what is a scalar. Since $T^\prime$ is an orthogonal set by induction, most pairs of elements in $T$ are already known to be orthogonal. We just need to test “new” inner products, between $\vect{u}_p$ and $\vect{u}_i$, for $1\leq i\leq p-1$. Here we go, using summation notation, \begin{align*} \innerproduct{\vect{u}_i}{\vect{u}_p}&= \innerproduct{\vect{u}_i}{ \vect{v}_p-\sum_{k=1}^{p-1}\frac{\innerproduct{\vect{u}_k}{\vect{v}_p}}{\innerproduct{\vect{u}_k}{\vect{u}_k}}\vect{u}_k } \\ &= \innerproduct{\vect{u}_i}{\vect{v}_p} - \innerproduct{\vect{u}_i}{ \sum_{k=1}^{p-1}\frac{\innerproduct{\vect{u}_k}{\vect{v}_p}}{\innerproduct{\vect{u}_k}{\vect{u}_k}}\vect{u}_k } &&\knowl{./knowls/theorem.IPVA.knowl}{\text{Theorem IPVA}}\\ &= \innerproduct{\vect{u}_i}{\vect{v}_p} - \sum_{k=1}^{p-1}\innerproduct{\vect{u}_i}{ \frac{\innerproduct{\vect{u}_k}{\vect{v}_p}}{\innerproduct{\vect{u}_k}{\vect{u}_k}}\vect{u}_k } &&\knowl{./knowls/theorem.IPVA.knowl}{\text{Theorem IPVA}}\\ &= \innerproduct{\vect{u}_i}{\vect{v}_p} - \sum_{k=1}^{p-1}\frac{\innerproduct{\vect{u}_k}{\vect{v}_p}}{\innerproduct{\vect{u}_k}{\vect{u}_k}}\innerproduct{\vect{u}_i}{\vect{u}_k} &&\knowl{./knowls/theorem.IPSM.knowl}{\text{Theorem IPSM}}\\ &= \innerproduct{\vect{u}_i}{\vect{v}_p} - \frac{\innerproduct{\vect{u}_i}{\vect{v}_p}}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\innerproduct{\vect{u}_i}{\vect{u}_i} - \sum_{k\neq i}\frac{\innerproduct{\vect{u}_k}{\vect{v}_p}}{\innerproduct{\vect{u}_k}{\vect{u}_k}}(0) &&\text{Induction Hypothesis}\\ &= \innerproduct{\vect{u}_i}{\vect{v}_p} - \innerproduct{\vect{u}_i}{\vect{v}_p} - \sum_{k\neq i}0\\ &=0 \end{align*}