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 the empty function, so that no Cauchy suite exists making our empty space complete. Gérard Lang.
closed as offtopic by abx, Goldstern, YCor, Ramiro de la Vega, Andrés E. Caicedo Sep 13 '18 at 19:47
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This is not true. Take a countable successor ordinal with the order topology. This topology makes it compact hence Polish.

$\begingroup$ All countable ordinals will do: they're all Polish. $\endgroup$ – Henno Brandsma Sep 13 '18 at 21:51
Take any countable, closed subset of a Polish space and it again will Polish. There are nondiscrete examples of this, like $\{\frac{1}{n}\}_{n=1}^\infty\cup\{0\}$ as a subset of $\mathbb{R}$.

1$\begingroup$ This is Tomek's example too: just $\omega+1$, the first countable successor ordinal. $\endgroup$ – Henno Brandsma Sep 13 '18 at 21:50