This may have a simple answer, but I couldn't find it so far either in textbooks or in math.stackexchange. Let $X$ be a metric space, and $$A=\bigcap^\infty_{n=1}A_n$$ a $G_\delta$ subset of $X$, where $A_n\subset X$ is open for each $n\in\mathbb{N}$. We assume for simplicity that $A_n\supset A_{n+1}$ for all $n\in\mathbb{N}$.
Question: Given any $B\subset X$ open such that $B\supset A$, is there an $n\in\mathbb{N}$ such that $B\supset A_n$?
On
No. Consider $\mathbb R$ with the usual metric. Take $A_n=(0,\frac{1}{n})$. Then $A=\emptyset$, yet there are plenty of $B$ that do not contain any of the $A_n$.
On
The answer is not necessarily, even if we assume that $X$ is complete. The reason is that $A$ itself might be open, and a proper subset of any of the $A_n$'s.
Consider $X=\Bbb N$ with the discrete metric, then let $A_n=\{0\}\cup\{k\in\Bbb N\mid k\geq n\}$.
We can also require that $X$ is connected, take $X=\Bbb R$ and let $A_n=(0,1)\cup(2,2+\frac1n)$.
Not necessarily.
Let $X=\Bbb R\setminus\{1\}$ under the subspace topology, $B=(0,1),$ and each $A_n=X\cap\left(0,1+\frac1n\right).$ Then $A=B,$ but no $A_n$ is a subset of $B$.
It may be true if we require the metric space to be complete, but I'm not sure. I'll think on it.
Edit: Drat! I see that Ittay already posted the super-nice counterexample I was about to add. And now Asaf has posted another (which hadn't occurred to me).
The upshot is that if the $G_\delta$ set $A$ is open, and is a proper subset of the $A_n$s, then letting $B=A$ gives a counterexample.
Second Edit: It occurs to me (belatedly, of course!), that my original example can be easily adapted into a counterexample where $X$ is complete, connected, and $A$ is non-empty.
Instead, put $X=\Bbb R$ in the usual (completely metrizable and connected) topology, $B=(0,1),$ and $A_n=(0,1)\cup\left(1,1+\frac1n\right).$ (Aside from $X$, these are all exactly the same sets.)