It's a well-known result that if X is a Hilbert space, then for any closed...

No. Counterexamples include, for example, $L^p$ spaces, where $p\in (1,2)\cup (2,\mathrm{\infty })$.

A normed linear space X that satisfies this property if and only if every non-empty, closed, convex subset of X is Chebyshev. That is, given any $x\in X$, and non-empty, closed, convex subset C of X, there is a unique point $c\in C$ such that
$\Vert x-c\Vert =d(x,C)=\underset{y\in C}{inf}\Vert x-y\Vert .$
Note that, taking x=0, this implies that the norm achieves a unique minimum over C.

Conversely, if X satisfies the property you want, and $C\subseteq X$ is non-empty, closed, and convex, then given $x\in X$, the norm should achieve a unique minimum on the set C−x; let this minimising point be y. Then, y=c−x for some $c\in C$, and
$\Vert c-x\Vert \le \Vert d-x\Vert$
for all $d\in C$, i.e. c is the unique closest point to x in C, i.e. C is Chebyshev.

Now, X has this property if and only if X is reflexive and strictly convex. In fact, the convex subsets of X will have at least one nearest point if and only if X is reflexive, and will have at most one nearest point if and only if X is strictly convex.

To see why this is, suppose X is reflexive. Then $B_X$ is weakly compact, as is any closed, bounded, convex set. We can express the set of nearest points of x to C as
$\bigcap _{\epsilon >0}B[x;d(x,C)+\epsilon ],$
where each term in the intersection is weakly compact, and hence the intersection is non-empty. That is, reflexivity implies the existence of nearest points.

Inversely, if X is not reflexive, then James' theorem guarantees us a bounded linear functional f that fails to achieve a maximum on the closed unit ball. It's not hard to see that 0 has no nearest point on $f^{-1}[1,\mathrm{\infty })$ (and in fact, no point outside the set has a nearest point inside the set).

If f is strictly convex, then the unit ball has no proper non-trivial faces, or equivalently, the only convex subsets of the unit sphere are the empty set and singletons. We can express the set of nearest points to x from a set C by
$B[x;d(x,C)]\cap C=S[x;d(x,C)]\cap C$
If C is convex, the left side is convex, but is a subset of the sphere, and hence must be a singleton or the empty set. That is, convex sets admit at most one nearest point.

Inversely, if X is not strictly convex, then we can find a hyperplane that supports the unit ball on at least a line segment. Then, for such a hyperplane, the origin projects onto at least that line segment, which means convex sets may admit more than one nearest point, completing the full proof.