I use the theorem that a closed subspace, Y, of a complete space, X, is complete; also that an intersection of open subsets of a separable space is separable. The latter is true because a subset of the countably dense subset of the Polish space, X, is dense in the intersection subspace, Y. What remains to be shown is that the intersection subspace, Y, is complete. The intersection subspace, Y, with the restricted metric is the complement of a union of closed complement subspaces. Each of these is complete by the aforementioned theorem. Their union is complete, because far enough out in any Cauchy sequence, the points lie in one of the complements . Finally, the complement of a complete subspace of a complete space is complete. QED? Is there a simpler proof?

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    $\begingroup$ Every subspace of a Polish space is the intersection of open sets, because every subspace is the union of closed sets, because every set is the union of singleton sets and singleton sets are closed. Did you perhaps mean to say "intersection of countably many open subsets"? $\endgroup$ – bof Jul 11 '17 at 0:38
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    $\begingroup$ Why do you believe that "the complement of a complete subspace of a complete space is complete"? $\endgroup$ – Andreas Blass Jul 11 '17 at 2:10
  • $\begingroup$ A Polish space space is second countable so all of its subspaces are second countable (so separable as well). You need to do dome work with the countably many open sets, you need a new metric in general: the irrationals are such an intersection in the reals, but their usual metric is far from complete. But it is possible (using the technique in my answer ) to define an alternative complete metric. $\endgroup$ – Henno Brandsma Jul 11 '17 at 8:52

If $(X,d)$ is a complete metric space, and $A = \cap_n U_n$ is a countable intersection of open subsets of $X$, then we can define a new metric $d'$ on $A$, equivalent to the inherited metric $d$ on $A$ that is also complete, namely:

$d'(x,y) = d(x,y) + \sum_n \frac{1}{2^n} f_n(x,y)$ where $f_n(x,y) = \min(\left|\frac{1}{d(x,X\setminus U_n)}-\frac{1}{d(y,X \setminus U_n)}\right|, 1)$.

The $\min$ guarantees that the series always converges (also due to the $\frac{1}{2^n}$ parts). $d \le d'$ and $d'$ is $d$-continuous they induce the same topology. A Cauchy sequence cannot go near $X\setminus A$, because this causes the $\frac{1}{d(x,X \setminus U_n}$-terms to "explode".

I'll leave the details to you.

You could also check that $i(x) = (x, (\frac{1}{d(x, X\setminus U_n)})_n)$ is an embedding of $A$ into $X \times \mathbb{R}^{\mathbb{N}}$, such that $i[A]$ is closed. The metric I just gave is the inherited metric from the complete space $X \times \mathbb{R}^{\mathbb{N}}$, pulled back to $A$.


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