Let $A$ be an inner product space of dimension $n$ with inner product $\langle\,,\,\rangle$.
Let $B$ be subspace of $A$, and let $\{a_1, ..., a_m\}$ be a basis for $B$.
Suppose that $\{a_1, ..., a_m, a_{m+1}, ..., a_n\}$ is a basis for $A$. Let $\{{a_1}', {a_2}', ..., {a_m}', ..., {a_n}'\}$ be the orthogonal basis for $A$ obtained by an application of the Gauss-Schmidt orthogonalization process on $\{a_1, a_2, ..., a_m, ..., a_n\}$.
Show that $\{a_{m+1}', ..., {a_n}'\}$ is a basis for $B^{\bot}$, and, $A = B + B^{\bot} $
Attempt:
Choose any orthonormal basis $E = \{e_1, ..., e_k\} \in B$.
Continue $E$ to a basis $E^{\bot} = \{e_1, ..., e_k; f_{k+1}, ..., f_n\}$ in $A$.
Applying GSOP to $E^{\bot}$. The first $k$ vectors in $E^{\bot}$ are orthogonal. So, GSOP alters nothing.
The remaining $n-k$ vectors may change, and we get the orthonormal basis $E^{\bot\bot} = \{e_1, ..., e_k; e_{k+1}, ..., e_n\}$ in $A$.
The vectors $e_{k+1}, ..., e_n$ are orthogonal to $e_1, ..., e_k$, and so they are all in $B^{\bot}$.
Present any $a\in A$ as the linear combination $a = (x_1e_1 + ... + x_ke_k) + (x_{k+1}e_{k+1} + ... + x_ne_n) = b + b'$
Hence, $b\in B$ and $b' \in B^{\bot}$. Which means that $\{a_{m+1}', ..., {a_n}'\}$ is a basis for $B^{\bot}$
Since $\{a_{m + 1}', \ldots, a_n'\}$ is orthonormal, it is also linearly independent (alternatively, since it is a subset of a linearly independent set). All you have to do is show that its span is $B^\perp$. To do this, let $b \in B^\perp$ and write it as a (unique) linear combination of the vectors in the orthonormal basis on $A$ (you should end up with a linear combination whose only nonzero coefficients are those for $a_{m + 1}', \ldots, a_n'$).
It seems that the key fact you're missing is that since $\{a_1', \ldots, a_n'\}$ is orthonormal, $$b = \langle b, a_1'\rangle a_1' + \cdots + \langle b, a_n'\rangle a_n'.$$ To finish the proof, you just use the definition of $B^\perp$ and the fact that $a_1', \ldots, a_m' \in B$.