Show that a countable dense subset $D \subset X$ is not a $G_{\delta}$

Question

Given $X$ a complete metric space with no isolated points and $D \subset X$ a countable dense subspace, show that $D$ is not a $G_{\delta}$.

I am quite lost in trying to use the hypothesis of the problem to complete the proof.

Since $X$ has no isolated points, for every $x \in X$, $\operatorname{int}(\{x\})=\emptyset$. I also know that $D$ is dense and countable, and I think I should use (maybe) Baire's Category Theorem. However, I am not being able to proceed. Any hint?

Answer 1

Suppose $D = \bigcap_{n\in \mathbb{N}} U_n$, where the $U_n$ are open sets. Then each $U_n$ is dense since it contains the dense set $D$.

Moreover we can write $X \backslash D = \bigcap_{x \in D} X \backslash \{x\}$, and each set $X \backslash \{x\}$ is open and dense since each $x \in D$ is not an isolated point.

Therefore $$\emptyset = D \cap (X \backslash D) = \bigg( \bigcap_{n \in \mathbb{N}}U_n \bigg)\cap\bigg( \bigcap_{x \in D} X \backslash \{x\} \bigg)$$ which is the countable intersection of open dense sets (since $D$ is countable).

But Baire says that a countable intersection of open dense subsets of $X$ cannot be empty, which contradicts the expression above.

Answer 2

Hint: The complement of $D$ is the countable union of nowhere dense sets.