If every metric on a set X is equivalent to the discrete metric then can we say X is a finite set? I know the result and the proof of the statement that
If X is a finite set then any two metrics are equivalent and every metric on X is equivalent to the discrete metric on X.
But I don't know whether the above result characterize finite sets?
 A: Assuming (a weak form of) the axiom of choice, yes, this is true.
Specifically, from the axiom of choice it follows that if $X$ is infinite then there is an injection $f:\mathbb{Q}\rightarrow X$ (of course this $f$ isn't unique in any sense). Consequently, assuming choice we have that every infinite set can be equipped with a metric so that it has a subspace isometric to $\mathbb{Q}$.
In a bit more detail, given such an $f$, we can define a non-discrete metric $d$ on $X$ as follows:

*

*If $a\not\in ran(f)$ then, for every $b\in X$ other than $a$ itself, we set $d(a,b)=d(b,a)=2$.


*Meanwhile, for $m,n\in\mathbb{Q}$ we set $d(f(m),f(n))=\min\{\vert m-n\vert, 1\}$.
Basically, $(X,d)$ looks like a "trunacted" copy of $\mathbb{Q}$ with a bunch of discrete points scattered around it, and is certainly not discrete!
(Note that this doesn't actually get a metric on $X$ with a subspace isometric to $\mathbb{Q}$; getting that is a good exercise.)

However, if we don't assume the axiom of choice, things can be weirder: it is consistent with $\mathsf{ZF}$ (= set theory without choice) that there are infinite sets which cannot be partitioned into two infinite subsets. Such sets, called amorphous sets, cannot support a non-discrete metric; this is a good exercise.
