Here is what I tried:
Since every set $G_n$ is open and dense, for any point $p$ in $G_1$, there is a neighborhood $N_{r}(p)$ which is completely contained in $G_1$ and contains infinitely many points of $G_n$ for all $n$ since $p$ is a limit point of all $G_n$ and so the intersection must be nonempty. I didn't use the fact that $X$ is complete and this looks too simple overall so there is probably something wrong with this reasoning.