Let $(X,d)$ be a complete metric space with no isolated points, and let $D\subset X$ be a countable dense set. Then $D$ is not a $G_\delta$ set.

Proof

First, we will prove that $D$ is first category. Then, supposing that $D$ is a $G_\delta$ we will conclude that $X - D$ must be first category. But then so must be $X$ which is absurd because $X$ is complete.

1) $D$ is a first category set:

By hypothesis $D=\left\{ {x_i } \right\}_{i\in N}$ Let's see that each singleton is nowhere dense if $X$ has no isolated points:

$\overline {\left\{ {x_i } \right\}} =\left\{ {x_i } \right\}$ (trivially). Suppose that $\left\{{x_i } \right\}^o=\left\{ {x_i } \right\}$ Then there is a ball aisolating the point. Absurd ($X$ has no isolated points). Then $\left\{ {x_i }\right\}^o=\emptyset$ and we have that $\left( {\overline {\left\{ {x_i } \right\}} }\right)^o=\emptyset$ so every singleton is nowhere dense and $D$ is of first category because it is a countable union of nowhere dense sets.

2) Suppose $D$ is a $G_\delta$ that is, $D=\bigcap\limits_{i=1}^\infty {U_i }$ such that every $U$ is open. As $D$ is dense, then each $U$ is dense, because $D\subset U_i \Rightarrow \overline D =X\subset \overline {U_i } \quad \forall i$ But then $X-D=X-\bigcap\limits_{i=1}^\infty {U_i } =\bigcup\limits_{i=1}^\infty {(X-U_i) }$ and $\left( {\overline {X-U_i } } \right)^o=\left( {X-U_i ^o} \right)^o=\left( {X-U_i } \right)^o=X-\overline {U_i } =\emptyset $ which implies that $X-D$ is of first category. Then $D\bigcup {(X-D)} = X $ is of first category. Absurd, because $X$ is complete.