Showing that metric induces single unique topology on a finite set

I am trying to prove, that given a metric on a finite set it induces exactly one topology. I have an idea which might lead to a proof, but am not sure:

For a finite set X with a given metric d we can prove it is a discrete topology:

$\forall x \in X$ take $r$ s.t. $r := \min_{y \in X}(d(x, y))$. We can do this as $X$ is finite. This way we can construct open balls for each $x$ such that they contain only x. So each x is in a open subset which is a singleton. And therefore we have a discrete topology.

I am not sure how to proceed showing that it is only this topology that we can get from the metric space.

Thank you.

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didn't you just prove that's the only topology you can get? – Dustan Levenstein May 6 '12 at 12:30
I do not know how to rigorously prove it is the only one. As from the question it seems that for infinite sets one can get many topologies using the same metric. – bim_bam May 6 '12 at 12:34
It's the other way round: There can be several metrics which induce the same topology. \\ Since you have started with an arbitrary metric on the set $X$, I would say your done with your proof. – Martin Sleziak May 6 '12 at 12:36
Your argument shows that any metric defined on a finite set will induce the discrete topology on that set. Therefore, the discrete topology is the only one that can be induced by a metric on a finite set. – John May 6 '12 at 13:56

Your proof is fine, except that you want your minimum to run only over $y\in X\setminus\{x\}$.
[Note that this proof (of course) fails for infinite $X$ as the minimum of infinitely many positive numbers need not exist (or, if we switch to the infimum, it need no longer be positive).]