Let $X$ be a topology space, $U\subseteq X$ is open set in $X$ and $A\subseteq X$ is dense in $X$
Prove: $\overline{U}=\overline{A\cap U}$
$\subseteq:$
Let $u\in U\subset \overline{U}$ be a non empty open set now $\overline{A}=X$ so $u\in\overline{A}\cap \overline{U}=\overline{A\cap U}$
It is correct? how should I attack $\supseteq$?
Best Answer
Since $A\cap U\subseteq U$, you can immediately derive that $\overline{A\cap U}\subseteq\overline{U}$, because it's a general fact that doesn't depend on $A$ being dense or $U$ being open.
The hypotheses are necessary to go the other way around (which is the part you attempted, but with no success, I'm afraid).
Let $x\in\overline{U}$. We need to show that $x\in\overline{A\cap U}$. Let $V$ be an open neighborhood of $x$. Then $U\cap V$ is an open set so, by density, it contains a point $a\in A$. Since $a\in U\cap V$, we have $a\in A\cap U$. Thus $a\in V\cap(A\cap U)$.