We proceed by induction on $n$. In the case $n=1$, we suppose $\{\mathbf{u_{k}}\}_{{k=1}}^{n}=\{\mathbf{u_{k}}\}_{{k=1}}^{1}=\mathbf{u_{1}}$ is a basis for the inner product space $V$. Letting $\mathbf{v_{1}}=\tfrac{\mathbf{u_{1}}}{\mid\mid\mathbf{u_{1}}\mid\mid}$, it is clear that $\mathbf{v_{1}}\in\spn\big(\mathbf{u_{1}}\big)$, whence it follows that $\spn\big(\mathbf{v_{1}}\big)=\spn\big(\mathbf{u_{1}}\big)=V$. Thus $\mathbf{v_{1}}$ is an orthonormal basis for $V$, and the result holds for $n=1$. Now let $n\geq 1\in\mathbb{N}$, and suppose the result holds for arbitrary $n$. Let $\{\mathbf{u_{k}}\}_{{k=1}}^{{n+1}}$ be a basis for an inner product space $V$. By the inductive hypothesis we may use $\{\mathbf{u_{k}}\}_{{k=1}}^{n}$ to construct an orthonormal set of vectors $\{\mathbf{v_{k}}\}_{{k=1}}^{n}$ such that $\spn\big(\{\mathbf{v_{k}}\}_{{k=1}}^{n}\big)=\spn\big(\{\mathbf{u_{k}}\}_{{k=1}}^{n}\big)$. In accordance with the procedure outlined in the statement of the theorem, let $\mathbf{m_{{n+1}}}$ be defined as

First we show that the vectors $\mathbf{v_{1}},\mathbf{v_{2}},\ldots,\mathbf{v_{n}},\mathbf{m_{{n+1}}}$ are mutually orthogonal. Consider the inner product of $\mathbf{m_{{n+1}}}$ with $\mathbf{v_{j}}$ for $1\leq j\leq n$. By construction, we have

Now since $\{\mathbf{v_{k}}\}_{{k=1}}^{n}$ is an orthonormal set of vectors, whence $\big<\mathbf{v_{i}},\mathbf{v_{j}}\big>=\delta_{{ij}}$, each term in the summation on the right-hand side of the preceding equation will vanish except for the term where $i=j$. Thus by this and the preceding equation, we have

Thus $\mathbf{m_{{n+1}}}$ is orthogonal to $\mathbf{v_{j}}$ for $1\leq j\leq n$, so we may take $\mathbf{v_{{n+1}}}=\tfrac{\mathbf{m_{{n+1}}}}{\mid\mid\mathbf{m_{{n+1}}}\mid\mid}$ to have $\{\mathbf{v_{k}}\}_{{k=1}}^{{n+1}}$ an orthonormal set of vectors. Finally we show that $\{\mathbf{v_{k}}\}_{{k=1}}^{{n+1}}$ is a basis for $V$. By construction, each $\mathbf{v_{k}}$ is a linear combination of the vectors $\{\mathbf{u_{k}}\}_{{k=1}}^{{n+1}}$, so we have $n+1$ orthogonal, hence linearly independent vectors in the $n+1$ dimensional space $V$, from which it follows that $\{\mathbf{v_{k}}\}_{{k=1}}^{{n+1}}$ is a basis for $V$. Thus the result holds for $n+1$, and by the principle of induction, for all $n$. ∎