Let $X$ be a nonempty perfect Polish space and let $Q$ be a countable dense subset of $X$. Then $Q$ is $F_{\sigma}$ but not $G_{\delta}$.
My question is about the $G_{\delta}$ part; specifically, I am unsure why the hypothesis that $X$ is perfect is included. Below is my (attempted) proof of why $Q$ is not $G_{\delta}$.
To show $Q$ is not $G_{\delta}$, suppose that it is. So then $Q$ is Polish, because it is a $G_{\delta}$ subset of a Polish space. Then $Q$ is also a Baire space, because it is completely metrizable. Baire spaces are not meager, because they are themselves open. However, $Q$ is meager, because it is $\bigcup_{q \in Q} \{ q \}$ (countable union because $Q$ is countable), where each singleton is nowhere dense. But $Q$ cannot be both meager and nonmeager, so the assumption that $Q$ is $G_{\delta}$ causes a contradiction and is therefore false.
Edit: It seems I need the perfectness to show that each singleton $\{q\}$ is actually nowhere dense (if $q$ were an isolated point, it wouldn't be nowhere dense) so that $Q$ is meager.
Best Answer
Bear in mind that the discrete topology on $\mathbb{N}$ is Polish (use a metric where each pair of distinct points is at a distance $1$ from each other). In this space, each singleton is an open set.
The mistaken step in your proof is where you say that each singleton must be nowhere dense (and this is why the statement included the assumption that $X$ is perfect).