First consider the following theorem:
Let $H$ be a hilbert space and $S$ be a closed subspace of $H$. $H=S \oplus S^{\perp}$, i.e. for $x \in H$ there exists (only one) representation of $x$ as $x=s+y $ for $s \in S$ and $ y \in S^{\perp}$.
Using the above Theorem I want to prove that: Let $P_S$ denote the orthogonal projection. For $x,y \in H$ holds $x \in S $ , $x-y \perp S \Leftrightarrow x=P_S y$.
$\Rightarrow$
Assume that $x \in S$ and $x-y \perp S$ holds. Take $h \in H$, then $h$ can be written as $h=(x-y)+s$ for some $s \in S$
Since $(x-y) \in S^{\perp}$ there exists some $s_0$ in $S$, such that $P_Ss_0=x-y$. I am not really sure how to further argue.
$\Leftarrow$ Assume $x = P_Sy$, then for some $s \in S$ $\langle x-y,s \rangle= \langle x,s \rangle-\underbrace{\langle y,s \rangle}_{=0 \text{ since } y \in S^{\perp}}=\langle x,s \rangle$
$\langle x,s \rangle$ now is $=0$ if $x \in S^{\perp}$, but what I want is to show that $x \in S$. How do I do that?
Note: I do know how to show the wanted result, but in this prove the theorem mentioned is not used. I heard that one can do this proof by using said theorem. And since I am new to hilbert projection theorem, I wanted to do some exercise to get a feeling for it.
Edit: I think I found the solution for the "$\Rightarrow$" direction: If $x \in S$ and $x-y \in S^{\perp}$, then for every $s \in S$ we have $x-s \in S$, because $S$ is a subspace. Now we get $x-y \perp x-s$. Using pythagoras we get $\lVert y-s \rVert^2=\lVert x-y \rVert^2+\lVert x-s \rVert^2 \geq \lVert x-y \rVert$. since $s$ was arbitrary, $\lVert x-y \rVert=min\{\lVert x-s \rVert : s \in S \}$, thus $P_Sy=x$