No. Counterexamples include, for example, spaces, where .
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 , and non-empty, closed, convex subset C of X, there is a unique point such that
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 is non-empty, closed, and convex, then given , the norm should achieve a unique minimum on the set C−x; let this minimising point be y. Then, y=c−x for some , and
for all , 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 is weakly compact, as is any closed, bounded, convex set. We can express the set of nearest points of x to C as
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 (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
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.