It's a well-known result that if X is a Hilbert space, then for any closed convex subset C of X, there exists a unique element of C with minimal norm. I'm wondering whether the converse is true. Let X be a Banach space such that for any closed convex subset C of X, there exists a unique element of C with minimal norm. My question is, does this imply that the norm on X is induced by some inner product? Now a norm is induced by some inner product if and only if it satisfies the parallelogram law. So does this property about closed convex sets somehow imply the parallelogram law? If not, does anyone know of a counterexample?

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

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 $x\in <!--\; \in \; -->X$, and non-empty, closed, convex subset C of X, there is a unique point $c\in <!--\; \in \; -->C$ such that
$\Vert <!--\; \Vert \; -->x-<!--\; -\; -->c\Vert <!--\; \Vert \; -->=d(x,C)=\underset{y\in <!--\; \in \; -->C}{inf}\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y\Vert <!--\; \Vert \; -->.$
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 $C\subseteq <!--\; \subseteq \; -->X$ is non-empty, closed, and convex, then given $x\in <!--\; \in \; -->X$, the norm should achieve a unique minimum on the set C−x; let this minimising point be y. Then, y=c−x for some $c\in <!--\; \in \; -->C$, and
$\Vert <!--\; \Vert \; -->c-<!--\; -\; -->x\Vert <!--\; \Vert \; -->\le <!--\; \le \; -->\Vert <!--\; \Vert \; -->d-<!--\; -\; -->x\Vert <!--\; \Vert \; -->$
for all $d\in <!--\; \in \; -->C$, 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
$\underset{\mathrm{\epsilon <!--\; \epsilon \; -->}>0}{\bigcap <!--\; \bigcap \; -->}B[x;d(x,C)+\mathrm{\epsilon <!--\; \epsilon \; -->}],$
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 $f^{-<!--\; -\; -->1}[1,\mathrm{\infty <!--\; \infty \; -->})$ (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
$B[x;d(x,C)]\cap <!--\; \cap \; -->C=S[x;d(x,C)]\cap <!--\; \cap \; -->C$
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.