It appears that this is not true if H is of infinite dimension. My question is therefore the following: does anyone have a counterexample? Is there a caracterisation for the points of the boundary that are (non trivial) projections? Thanks in advance.

I'm a little concerned that my original incorrect answer was accepted and the only feedback on the edit was a comment that it didn't make sense ... here's a slightly simpler counterexample. Let $H = l^2$ and let $C$ be the set of sequences $(a_n)$ satisfying $a_n \leq \frac{1}{n}$ for all $n$. $C$ is clearly closed and convex. $0$ is a boundary point of $C$ (in fact $C$ has no interior), but it is not the nearest element of $C$ to any point outside of $C$. If $(a_n)$ is any nonzero sequence in $l^2$ then some entry $a_{n_0}$ must be nonzero, and then we can find a point in $C$ that is closer to $(a_n)$ than $0$ is. For sufficiently small $\epsilon$, the point $\epsilon a_{n_0}e_{n_0}$ (where $e_n$ is the standard basis) works. Another way to say this is that $0$ has no support hyperplane. 

