Let $V$ be a normed vector space. It is said to be strictly convex if its unit sphere does not contain nontrivial segments. A subset $A \subset V$ is said to have the unicity property if for any $x \in V$, there is exactly one $x' \in A$ with $|x - x'| = \inf_{y \in A}|x - y|$.
If $V$ is strictly convex then any finite dimensional linear subspace of it has the unicity property.
Does the converse hold?
Yes, the converse is also true. Suppose the space is not strictly convex. Let $[a,b]$ be a line segment contained in the unit sphere. The function $$t\mapsto \|(1-t)a+tb\|,\qquad t\in\mathbb R\tag1$$ is convex and is equal to $1$ on $[0,1]$. Therefore, it is greater than or equal to $1$ everywhere.
The distance from $0$ to the line (1) is realized by any point of $[a,b]$. Apply translation by $-a$ to conclude that the distance from $-a$ to the line $t\mapsto t(b-a)$ is realized by multiple points. Therefore, $V$ does not have the unicity property.