$X$ a separable metric space with no isolated points. If $G \subset X$ is a countable dense $G_{\delta}$ subset of $X$, why is $X$ meager?

Question

Let $X$ be a separable metric space with no isolated points. Then if $G \subset X$ is a countable dense $G_{\delta}$ subset of $X$, why is $X$ meager?

I've written a couple things so far,

Let $G=\bigcap U_n$, where each $U_n$ is open in $X$ and $G$ is countable and dense. It follows that each $U_n$ is dense in $X$. Also, $G^c=\bigcup U_n^c$, a union of closed sets.

Answer 1

Let $G=\{x_n:n\in\Bbb N\}$, and for $n\in\Bbb N$ let $V_n=U_n\setminus\{x_n\}$; then each $V_n$ is still dense in $X$ (why?), and $\bigcap_{n\in\Bbb N}V_n=\varnothing$. Can you finish it from there?