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Actually,

I'm a little concerned that my first original incorrect answer was incorrectaccepted and the only feedback on the edit was a comment that it didn't make sense .Here's .. here's a slightly simpler counterexample.

Let $H = l^2$ and let $C$ be the closed convex hull set of the vectors $ae_n$ with sequences $n \in {\bf N}$ and (a_n)$ satisfying $|a| |a_n| \leq \frac{1}{n}$. Then frac{1}{n}$ for all $n$. $C$ is clearly closed and convex.

$0$ is a boundary point of $C$ --- for any $\epsilon > 0$ we can find $n > \frac{2}{\epsilon}$, and then $\frac{\epsilon}{2}e_n$ is not (in fact $C$ has no interior), but is in the $\epsilon$-ball about $0$. But $0$ it is not the projection onto nearest element of $C$ of to any vector point outside of $C$, because if C$. If $(a_n)$ is any nonzero element of sequence in $l^2$ it must have then some nonzero entry , say $a_n$, a_{n_0}$ must be nonzero, and then it we can find a point in $C$ that is closer to $ae_n$ for some nonzero $a$ (a_n)$ than it $0$ 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 $0$.0$ has no support hyperplane.
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Well

Actually, I think it's true even in infinite dimensionsmy first answer was incorrect. Assume real scalars, Here's a counterexample. Let $H = l^2$ and let $C$ be a the closed convex set in a Hilbert space hull of the vectors $H$ ae_n$ with $n \in {\bf N}$ and let $x$ be |a| \leq \frac{1}{n}$. Then $0$ is a boundary point . By shifting we can assume of $x = C$ --- for any $\epsilon > 0$ . Then since every closed convex set is an intersection of half-spaces, we can find $y \in H$ such that $\langle y,z\rangle n > \leq 0$ for all frac{2}{\epsilon}$, and then $z \\frac{\epsilon}{2}e_n$ is not in $C$ , and but is in the projection of $y$ \epsilon$-ball about $0$. But $0$ is not the projection onto $C$ equals 0 by a simple computation. The same conclusion holds of any vector outside of $C$, because if $(a_n)$ is any nonzero element of $l^2$ it must have some nonzero entry, say $a_n$, and then it is closer to $ae_n$ for complex scalars since every complex Hilbert space some nonzero $a$ than it is isometric to a real Hilbert space.$0$.
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Well, I think it's true even in infinite dimensions. Assume real scalars, let $C$ be a closed convex set in a Hilbert space $H$ and let $x$ be a boundary point. By shifting we can assume $x = 0$. Then since every closed convex set is an intersection of half-spaces, we can find $y \in H$ such that $\langle y,z\rangle \leq 0$ for all $z \in C$, and the projection of $y$ onto $C$ equals 0 by a simple computation. The same conclusion holds for complex scalars since every complex Hilbert space is isometric to a real Hilbert space.