A $G_\delta$ set is a countable intersection of open sets. The set of rational numbers is not a $G_\delta$ set.
But I'm wondering why the following construction doesn't make it a $G_\delta$ set. Let $(q_n)$ be a sequence enumerating the rational numbers. For any natural number $m$, let $U_m = \cup_n (q_n-\frac{1}{2^{m+n}},q_n+\frac{1}{2^{m+n}})$, and let $U = \cap_m U_m$. Then it seems to me that $U_m$ is an open set for all $m$, and $U$ is the set of rational numbers. So the set of rational numbers is a $G_\delta$ set.
Where is the flaw in my reasoning?
There are many ways to enumerate the rational numbers. Since you did not specify an enumeration, let me pick one for you.
For each positive integer $m$, choose a rational number $q_{2m}$ so that $|q_{2m}-\sqrt2|\lt\frac1{2^{3m}}$. In this way we get a sequence $q_2,q_4,q_6,\dots$ of rational numbers converging to $\sqrt2$. Enumerate the rest of the rational numbers any way you like in a sequence $q_1,q_3,q_5,\dots$. Conbining these, we get an enumeration $q_1,q_2,q_3,\dots$ of the set of all rational numbers.
Note that, for each positive integer $m$, we now have $$\sqrt2\in\left(q_{2m}-\frac1{2^{m+2m}},\ q_{2m}+\frac1{2^{m+2m}}\right)\subseteq\bigcup_{n=1}^\infty\left(q_n-\frac1{2^{m+n}},\ q_n+\frac1{2^{m+n}}\right)=U_m,$$ so $\sqrt2\in\bigcap_{m=1}^\infty U_m$, although $\sqrt2$ is not a rational number.