Finding the size $k$ subset in a metric space that maximizes the min distance between elements
From a quick Google search, it looks like your problem is sometimes called (metric) "facility dispersion." This paper by Ravi, Rosenkrantz, and Tayi seems to prove that your heuristic is a $2$-approximation, and that this factor is actually optimal by a reduction to clique.

I skimmed it, so I might be missing some subtle points, but the idea of the proof seems quite simple: take an optimal $S\subseteq X$ with $\vert S\vert=k$, with objective value $\mathsf{OPT}$. Note that the $k$th point you add in the greedy algorithm will attain the closest distance (to another point) in the greedy solution, so it suffices to show that the last point you add must have distance at least $\mathsf{OPT}/2$ from the previous $k-1$ points you selected. This is seen by noting that the $k$ (open) metric balls of radius $\mathsf{OPT}/2$ around each point in $S$ are disjoint; as a result, at least one of these balls does not contain any of the first $k-1$ points you have chosen. The respective center in $S$ is therefore an option at step $k$ which attains a pairwise distance of at least $\mathsf{OPT}/2$ from all previously chosen points, so the same must be true of the greedy point chosen at the $k$th step.