Is the closure of a countable $G_\delta$ set countable?

It's not true.

Let me think of $C = 2^\omega$ as the Cantor set in $[0,1]$. It's homeomorphic to its square, so I'll actually work in $C^2 \subset [0,1]^2$.

Let $E_n$ be the endpoints of all the intervals remaining at the $n$th step of the construction of $C$, so $|E_n| = 2^{n+1}$. Let $X = \bigcup_n E_n \times \{1/3^n\}$ which is countable, and indeed, discrete.

Now for every $x \in E = \bigcup_n E_n$, clearly $(x,0) \in \overline{X}$. But $E$ is dense in $C$ so we have $C \times \{0\} \subset \overline{X}$, and that's uncountable.

Indeed, we can see that $\overline{X} = X \cup C \times\{0\}$. So back in $C^2$, we can write $X = \overline{X} \cap (C \times \{0\})^c$, which is the intersection of a closed (hence $G_\delta$) set and an open set. Thus $X$ is $G_\delta$ in $C^2$.