What went wrong in this proof? 
Let $U \subset V$ and $P_{U}v$ to be an orthogonal projection of $V$ onto $U$. Show $\| P_Uv\| \leq \|v \|$

This geometrically makes sense. 
Here is what I wrote.
Let $(e_1, \dots,e_m, e_{m+1}, \dots, e_n)$ be an orthonormal basis for $V$. Then for any $v \in V$, we have $$v = \langle v,e_1 \rangle e_1 + \dots+ \langle v,e_n \rangle e_n$$ 
and
$$\| v \|^2 = \| \langle v,e_1 \rangle \|^2 + \dots + \| \langle v,e_n \rangle \|^2$$
Also, $P_U(v) = \langle v,e_1 \rangle e_1 + \dots + \langle v,e_m \rangle e_m$ where we let
$(e_1, \dots,e_m)$ be an orthonormal basis for $U$ and $(e_{m+1}, \dots, e_n)$ be orthonormal base for $U^\perp$.
Then we have 
$$\| v \|^2 = \| \langle v,e_1 \rangle \|^2 + \dots + \| \langle v,e_m \rangle \|^2 +  \underbrace{\| \langle v,e_{m+1} \rangle \|^2 + \dots + \| \langle v,e_n \rangle \|^2}_{\delta^2} = \| P_U v \|^2 + \delta^2$$
Rearrange a bit $\| P_U v \|^2 = \| v \|^2 - \delta^2  \leq \| v \|^2$, which gives the desired result
 A: There's, I think, a wrong understanding of things here and this may be a rather serious mistake:
You begin by choosing an orthonormal basis $\,\{e_i\}_{i=1}^n\;$ of the whole space $\,V\,$ and then you imply that from this very same basis you can choose a(n orthonormal) basis $\;\{e_i\}_{i=1}^m\;$ for $\;U\;$ and also for its complement. This is false and if you give it a little thought you can find an easy counterexample with $\,\Bbb R^2\;$ and the usual, canonical basis (and the euclidean inner product, of course.)
You could, and imo should, have gone the other way: first choose a basis for $\,U\;$, orthonormalize it by Gram-Schmidt, complete it to a basis of the whole $\,V\;$ and again use GS to orthonormalize and etc.
A: The thing wrong with your proof is that you don't introduce the basis in the correct way. The first time you mention it, it is just any orthonormal basis; you cannot in the sequel suddenly suppose it has any particular relation to the subspaces $U$ and $U^\perp$. But if with a bit of foresight, you start choosing an orthonormal basis $\{e_1,\ldots,e_m\}$ of the subspace $U$ (where $m=\dim U$) and another orthonormal basis $\{e_{m+1},\ldots,e_n\}$ of the subspace $U^\perp$, then you will have no difficulty in proving that $\{e_1,\ldots,e_n\}$ is an orthonormal basis of$~V$ (but not just any orthonormal basis), and using that basis your proof runs fine.
By the way, you don't really need to use any basis at all for this result. By the definition of orthogonal projection one has $P_U(v)\perp v-P_U(v)$, so that by Pythagoras' theorem $\|v\|^2=\|P_U(v)\|^2+\|v-P_U(v)\|^2$ whence $\|v\|\geq\|P_U(v)\|$.
