I am looking for an elegant proof of the fact that a countable metric space is complete iff its underlying topology is discrete.


It is easy to see that a discrete space is complete because its topology can be derived from the distance $d(x,y)=1$ iff $x$ and $y$ are distinct, so that every Cauchy sequence must be eventually constant, and converge to this constant point that is inside the space. But the proof in the other side, that the topology underlying a countable and complete metric space must be discrete does not seem so easy. 


By the way, I take it that:


(i) If the space is a singleton, $d(x,x)=0$ is a distance whose underlying topology is the discrete (and unique) topology on $X$, so that the only (Cauchy) sequence on x is constant and convergent to x, making the space complete;


(ii) If the space is empty, it is a countable space whose only topology can be considered as discrete. We can also consider that the only function from the empty set into the positive reals (and similarly into the natural numbers) is the empty function, so that no Cauchy sequence exists making our empty space complete.
Gérard Lang. 



Must a countable Polish space be discrete?
This is not true. Take a countable successor ordinal with the order topology. This topology makes it compact hence Polish.


