Let $(X,d)$ be a metric space and suppose there is at least one finite collection $\mathcal{S}$ of closed unit disks (with centers in $X$, though the disks need not be entirely in $\mathcal{S}$) which covers $X$ (that is, for all $x \in X$ there is a disk from $\mathcal{S}$ containing $x$).
If $\mathcal{S}_1$ and $\mathcal{S}_2$ are collections with the above property and of minimal size, then must there exist a bijection $f$ from the centers of the disks in $\mathcal{S}_1$ to the centers of the disks in $\mathcal{S}_2$ such that $d(x,f(x)) < 2$ for any center $x$ of a disk in $\mathcal{S}_1$? Does this change if $X$ is compact?
Note that if we replace the strict inequality condition by $d(x, f(x)) \leq 2$, then I found that the answer is "yes" (no matter whether we have compactness or not, I think) and a possible proof is by induction on the common size of $\mathcal{S}_1$ and $\mathcal{S}_2$. I highly expect that there is a silly counterexample (perhaps not in a Euclidean space) to the case I am asking about, though I cannot find one.
Any help appreciated!
There is a graph counterexample. Let $X$ be a cycle $v_1–\dots–v_7– v_1$ and for each vertices $v,u$ of $X$ let $d(v,u)$ be length of the shortest path from $x$ to $y$. Let $\mathcal S_1$ be a family of unit disks centered at $v_1$, $v_2$, and $v_5$ and $\mathcal S_2$ be a family of unit disks centered at $v_1$, $v_4$, and $v_5$.