I was asked to proof something in an exercise, and in the second half I would have to come up with a counterexample to a weaker statement.
I had to proof that given a topological space $(X, T_{X})$ and open, dense subsets $Y, Z \subset X$ then also $Y \cap Z$ is closed.
I tried to give a proof but I haven't used the fact that $Y, Z$ are open (or at least not explicitly). Therefore there must be some error in it I supppose, my proof is as follows: Well first, we know that $\bar{Z}=X, \bar{Y} = X$ so therefore we also know that $\bar{Y}\cap\bar{Z} = X$. We find that $\bar{Y} \cap \bar{Z}$ is closed and that we've got the following nested chain of subsets: $Y \cap Z \subset \bar{Y} \cap \bar{Z} \subset \overline{Y \cap Z}$ with the last inclusion following form the fact dat $\overline{Y\cap Z}$ is the greatest closed set which contains $Y \cap Z$. The first inclusion should be a simple set theoretic verification. ( Suppose $p \in Y \cap Z$ then $p \in Y, p \in Z$ so therefore $p \in \bar{Y}, p \in \bar{Z}$. we conclude $p \in \bar{Y} \cap \bar{Z}$.) Now because $\bar{Y} \cap \bar{Z} = X$ we conclude that it also must hold that $\overline{Y \cap Z} = X$
Where did I go wrong? Because we can also look at the following counterexample showing that we really do need opennes of $Y, Z$: Let $X$ be the reals with the euclidean topology and consider $Y = \mathbb{Q}, Z = \mathbb{R} \setminus \mathbb{Q}$. These sets are dense and closed, their intersection however is empty so it is not dense.
Suppose that $O$ is open and non-empty; we want to show it intersects $Y \cap Z$. As $Y$ is open, so is $Y \cap O$, and it's non-empty as $Y$ is dense. So it intersects the dense set $Z$, and so $\emptyset \neq (Y \cap O) \cap Z = O \cap (Y \cap Z)$ as required. Note we need only one of the dense sets to be open.