The exercise says
Let $(G_k)$ a sequence of open and dense subsets in a complete metric space. Show that $\bigcap_{k\geqslant 0}G_k$ is also open and dense
I have shown that $G:=\bigcap_{k\geqslant 0}G_k$ is dense, however I think that I have a counterexample for $G$ to be also open. If it would be true then it must also be true that $G^\complement =\bigcup_{k\geqslant 0}G_k^\complement $ would be closed and with empty interior. However setting $G_k^\complement :=\{1/k\}$ we have that $G^\complement=\{1/k:k\in \Bbb N_{> 0} \} $ have empty interior in $\Bbb R$ but it is not closed because zero is a limit point of $G^\complement $.
Im wrong or the exercise is wrong?
The exercise is wrong. The intersection of countably many open dense sets is dense, as you know, but doesn't have to be open. It is by definition $G_\delta$, the intersection of countably many open sets. Another example is if $G_k=\mathbb{R}\setminus\{{r_k}\}$ where $r_1,r_2,\ldots$ is an enumeration of the rationals. Then the intersection of the $G_k$ is the irrational numbers, which is definitely dense and not open.