In the case of the $\Bbb R$, $G_\delta$ subsets do not have many restrictions in terms of interior or denseness. But you can prove that every closed subset of $\Bbb R$ is also $G_\delta$. An equivalent, perhaps easier, thing to prove is that every open subset of $\Bbb R$ is $F_\sigma$. You can certainly prove that an open interval is $F_\sigma$. If you know in addition that every open subset of $\Bbb R$ is a countable disjoint union of open intervals and that a countable union of $F_\sigma$ subsets is still $F_\sigma$, you can prove that result.
The best way to see that $\Bbb Q$ is not a $G_\delta$ will often use the Baire Category Theorem. If you know that $\Bbb R$ is a Baire space, then you can show $\Bbb Q$ is not $G_\delta$ easily with this result that you should do as an exercise: A dense $G_\delta$ subset in any topological space is comeager. Why does this finish it? Because $\Bbb Q$ is already meager as it is a countable union of singletons. If $\Bbb Q$ were comeager, $\Bbb R$ would be the union of two meager subsets.