Show that the projection moves you closer to the set, i.e. $|P_C(x) - y| \leq |x-y|$ for all $y\in C$ Let $H$ be a Hilbert space with induced norm $|\cdot|$, with $C\subset H$ a closed, convex subset of $H$ and $P_C(x)$ the projection of $x$ onto $C$, i.e.,

$$P_C(x) = \arg\min\limits_{y\in C}|x-y|^2$$

I want to show the following trivial result,

$$\forall y\in C, \quad |P_C(x) - y|^2\leq |x-y|^2$$

I have tried a couple things but none have worked. The first was the following (let $y\in C, x\not\in C$):

$$|P_C(x) -y|^2 = |P_C(x)|^2 - 2\langle P_C(x), y\rangle +|y|^2 = |P_C(x)|^2 - 2\langle x, y\rangle +|y|^2 \\\leq |P_C(x)|^2 + |x-y|^2$$

I also tried this:

$$|P_C(x) - y|\leq |P_C(x) - x| + |x-y|\leq 2|y-x|$$

but the $2$ is ruining things for me. Can someone give me a hint?

Edit: here is what I came up with ultimately,

$$|x-y|^2 = |x-P_C(x) +P_C(x) - y|^2\\ = |x-P_C(x)|^2 -2\langle x-P_C(x), y-P_C(x)\rangle + |y-P_C(x)|^2\\
\geq -2\langle x-P_C(x), y-P_C(x)\rangle + |y-P_C(x)|^2\\
\geq |y-P_C(x)|^2$$

since the inner product is negative. Of course then we should actually show that the inner product is negative in general and this is basically the argument that bananach was referring to, I think.
Split $x-y$ into the path from $x$ through $P_C(x)$ and then in the orthogonal direction after some while. Then use Pythagoras. Let me know if you need details, but try just sketching this down on a piece of paper, it should become obvious.

Algebraically speaking, add and subtract $Px$ in |x-y|^2 and then use Pythagoras. The scalar product that appears is positive because $f(t)= |x-Px+t(y-Px)|^2$ is non decreasing in $t=0$ by definition of $Px$.
As $P_C(y)=y$ for $y\in C$ we know that $P_C(x)-y=P_C(x-y)$.  From here show for all $x\in H$ that $\|P(x)\|\leq \|x\|$.  For that purpose it's enough to prove
$$\|x\|^2=\|P_C(x)\|^2+\|x-P_C(x)\|^2.$$
Now that follows easily from $\langle x-P_C(x),P_C(x)\rangle=0$.