Union of locally finite family of $G_\delta$ sets Is an arbitrary union of a locally finite family of $G_\delta$ sets is again $G_\delta$?
Suppose my collection of $G_\delta$ sets are $\left\{G_\lambda=\bigcap_{i\ge 1} G_i^\lambda:\lambda\in\Lambda\right\}$, where $G_i^\lambda$ are open sets. Then, $$\bigcup_{\lambda\in\Lambda} G_\lambda=\bigcup_{\lambda\in\Lambda}\bigcap_{i\ge 1}G_i^\lambda=\bigcap_{\alpha\in\mathbb{N}^\Lambda} \bigcup_{\lambda\in\Lambda}G_{\alpha(\lambda)}^\lambda$$ Now, $A_\alpha=\bigcup_{\lambda\in\Lambda}G_{\alpha(\lambda)}^\lambda$ is clearly open for each $\alpha \in \mathbb{N}^\Lambda$. But I am unable to show that the intersection $\bigcap_{\alpha \in \mathbb{N}^\Lambda}A_\alpha$ can be made countable.
Is my approach correct? Also if the claim is false itself, can someone please provide any counter-example? Any help regarding this is appreciated!
 A: Define a new topology $\tau$ on $\Bbb R$ as follows. Each $x\in\Bbb R\setminus\Bbb Q$ is isolated. For $q\in\Bbb Q$ and $\epsilon>0$ let 
$$B(q,\epsilon)=\{q\}\cup\{x\in\Bbb R\setminus\Bbb Q:|x-q|<\epsilon\}\;;$$
$\{B(q,\epsilon):\epsilon>0\}$ is local base at $q$. I’ll use $X$ to denote the reals with the topology $\tau$.
Clearly $\Bbb Q$ is a closed, discrete subset of $X$. Moreover, $X$ is first countable, so $\{q\}$ is a $G_\delta$ for each $q\in\Bbb Q$, and $\big\{\{q\}:q\in\Bbb Q\big\}$ is a locally finite family of $G_\delta$ sets. 
Suppose that $\Bbb Q\subseteq U\in\tau$; then $U$ contains a Euclidean open nbhd of $\Bbb Q$, which is of course a dense open set in $\Bbb R$ with the usual topology. Let $\{U_n:n\in\Bbb N\}\subseteq\tau$ be a family of open nbhds of $\Bbb Q$ in $X$; without loss of generality we may assume that each $U_n$ is open in the usual topology. By the Baire category theorem $\Bbb Q$ is not a $G_\delta$ in the usual topology of $\Bbb R$, so $\bigcap_{n\in\Bbb N}U_n\ne\Bbb Q$, and $\Bbb Q$ is not a $G_\delta$ in $X$, either.
