The Hilbert space projetion theorem is the following theorem:
Let $H$ be a Hilbert space and $C$ any closed convex subset. Then for $h \in H$ there exists a unique $c_0 \in C$ such that $\|h-c_0\| = \inf_{c\in C}\|h-c\|$.
Please could someone show me how to finish my idea for a proof?
Proof:
Note that $C$ is a closed set therefore $\inf_{c\in C}\|h-c\| = \min_{c\in C}\|h-c\|$, or in other words: there exists $c_0 \in C$ with $\|h-c_0\|=\min_{c\in C}\|h-c\|$. This shows existence.
We have to show uniqueness next. Let $c_0' \in C$ be another element with $\|h-c_0'\| = \min_{c\in C}\|h-c\|$.
We have not yet used that $C$ is convex and that $H$ is a Hilbert space. Convexity could mean we have to use that ${1\over 2}(c_0 + c_0')\in C$ but I can't proceed from there. That $H$ is a Hilbert space could mean we have to use the parallelogram identity. I tried to apply it to $\|c_0 - c_0'\|^2 = \|(c_0 - h) - (c_0'-h)\|^2$ to show it is zero but couldn't finish the idea either.
I alluded to a method in my comment, but I guess I'll leave it here as an answer. Let $\delta=\inf_{c\in C}\|c-h\|$. Suppose $x,y\in C$ are such that $\|x-h\|=\|y-h\|=\delta$. Using the parallelogram identity, we have $$\|x-y\|^2=2\|x-h\|^2+2\|y-h\|^2-\|(x-h)+(y-h)\|^2=4\delta^2-4\|\frac{x+y}2-h\|^2.$$ Using the fact that $C$ is convex, $\frac{x+y}2\in C$ so $\|\frac{x+y}2-h\|\geq\delta$. From here we can deduce $\|x-y\|^2=0$.
Let $(x_n)$ be a sequence in $C$ with $\|x_n-h\|\to\delta$. Fix $\epsilon>0$. There exists $N>1$ such that if $n\geq N$, $\|x_n-h\|^2<\delta^2+\frac{\epsilon^2}2$. From here, you should be able to use similar steps to the above to show that $(x_n)$ is Cauchy. If $x=\lim_{n\to\infty}x_n$, it is quite simple to show $\|x-h\|=\delta$. I'll leave the last few details to you.