There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism While studying differential geometry I have formulated a result in linear algebra which I haven't been able to prove:

Let $V$ be a real vector space with an inner product, and let $\{e_1,\dots,e_k\}\subset V$ be an orthonormal basis. Let $W\subset V$ be a subspace of dimension $n\geq 1$. There exist $i_1,\dots,i_n$ such that $P_S|_W:W\to S$ is an isomorphism, where $S=\operatorname{span}(e_{i_1},\dots,e_{i_n})$ and $P_S:V\to S$ is the orthogonal projection onto $S$.

Here's (what I think is) the most fruitful approach I've tried.
I have geometrically visualized that "almost always" any $(i_1,\dots,i_n)$ will do, except when some $e_i\in W$. So, I assert that we can pick $(i_1,\dots,i_n)$ such that $W$ is not perpendicular to $\operatorname{span}(e_{i_j})$ for all $j=1,\dots,n$.
In that case, I try to prove that $P_S|_W$ is surjective. Let $v=\sum_j \langle v,e_{i_j}\rangle e_{i_j}\in S$. Define $w_{i_j}=P_W(e_{i_j})$ and $w=\sum_j \langle v,e_{i_j}\rangle w_{i_j}$. A computation gives $$P_S(w)=\sum_k \langle v,e_{i_k}\rangle P_S P_W(e_{i_k})$$
If I could prove that $P_S P_W(e_{i_k})\in \operatorname{span}(e_{i_k})$ (which I geometrically believe, although I have a feeling it might fail in higher-than-imaginable dimensions), then I could, redefining $w_{i_j}$ by a scalar multiple, get that $P_SP_W(e_{i_k})=e_{i_k}$ thus obtaining $P_S(w)=v$ and finishing the proof. But I haven't been able to obtain a useful expression of $P_SP_W(e_{i_k})$.
I'd appreciate any proof, but I'd especially appreciate any comments on this approach (I've been at it for some hours now).
 A: Induct on the dimension of $V$: The statement is clear when $n = 1$. First consider the projection $PW$ onto the span of $e_2,\ldots,e_n$. By induction, there are some $\{e_i\}_{i \in I}$ such that the projection $P'$ of $PW$ onto the span of the $e_i$ for $i \in I$ is an isomorphism. ($I \subseteq \{2,3,\ldots,n\}$). The kernel of $P$ when restricted to $W$ is either zero-dimensional or one-dimensional. In the first case we are done: $P'P$ is an isomorphism of $W$ onto the span of the $\{e_i\}$ for $i \in I$. In the second case, we must have $e_1 \in W$, and so we simply need to add $e_1$ to the list of vectors.
A: Well, for surjectiveness, we would need that $e_{i_k}\in P_S(W)$, because then any of their linear combination is also present. I tried to prove it, but, after rearranging the indices so that $S={\rm span}(e_1,..,e_n)$, it leads to ensuring that certain elements like $(1,0,0,..,\alpha_{n+1},\alpha_{n+2},..)$ are in $W$...
However, trying from the other end: we have $\ker P_S=S^\perp$, and we need $P_S|_W$ be injective, and $\dim S=\dim W$. Injectivity means  $\ker P_S|_W=W\cap S^\perp=\{0\}$.
So, we can choose $i_1$ so that $e_{i_1}\not\perp W$, this way, with $S_1:= {\rm span}(e_{i_1})$, we have  $(W\cap S_1^\perp)\subsetneq W$. 
Then, choose $i_2$ so that, with $S_2:={\rm span}(e_{i_1},e_{i_2})$, we have
$$W\cap S_2^\perp\,\subsetneq\, W\cap S_1^\perp\,,$$
and so on...
A: Update: this answer is wrong. See the comments below.

Ok, so I was riding the bus back home, a little obsessed by this problem, and I think I finally got around it:
We can suppose $n<k$, the theorem is obvious if $n=k$.
Now observe that $\operatorname{ker}(P_S|_W)=S^\perp \cap W$. So, we want to find indices $t_1,\dots,t_{k-n}$ such that $W\cap \operatorname{span}(e_{t_1},\dots,e_{t_{k-n}})=\{0\}$.
Assume by contradiction that for every choice of indices $t_1,\dots,t_{k-n}$ we have that $W\cap \operatorname{span}(e_{t_1},\dots,e_{t_{k-n}})\neq\{0\}$. Therefore $W$ contains some $e_i$ of each of these $\binom{k}{k-n}$ subspaces. A combinatorial argument (using $n<k$) shows that $W$ must have at least $k-(k-n)+1=n+1$ different $e_i$, which is absurd since $\dim W=n$.

I think the argument is nice, but I would appreciate how to really prove the "combinatorial argument" (I'm really bad at these).
A: Take a basis of $W$ and express its vectors in the basis $e_1,\ldots,e_k$, forming a $k\times n$ matrix with those columns (remember $n\leq k$). Since you had a basis this matrix has (full) rank $n$. So you can select $n$ linearly independent rows from the matrix, which means a subset of $n$ coordinates of those $n$ basis vectors that are linearly independent. Then the projection on the subspace spanned by the corresponding $n$ vectors among $e_1,\ldots,e_k$, parallel to the space spanned by the remaining ones, is an isomorphism of vector spaces (the projection matrix is formed by just those selected $n$ rows, and has nonzero determinant).
