Let $p, A, B$ be open sets such that $p \subset A$, $\bar A \subset \bar B$ and $p \cap A\neq \emptyset$. Does $p \cap B \neq \emptyset$ holds? I need this for a proof.

-
If $p\subset A$, then why do you write $p\cap A\neq \emptyset$? Isn't that redundant? – Stefan Hamcke Jun 16 '13 at 20:35
Just another way to say $p \neq \emptyset$ – Vinicius Rodrigues Jun 16 '13 at 20:37
But that's an unnecessarily complicated way to say that $p$ is non-empty, and I don't see why anyone in the world would possibly want to do that. – Stefan Hamcke Jun 16 '13 at 20:43
@DonAntonio: $\bar A\subset\bar B$ does not imply $A\subset B$. – Stefan Hamcke Jun 16 '13 at 22:56
I know, Stefan...I was sure I saw $\,\overline A\subset B\;$ but perhaps I simply misread. Deleting – DonAntonio Jun 16 '13 at 23:00

As $p \cap A$ is non-empty and open, and as $p \cap A \subset A \subset \overline{A} \subset \overline{B}$, it must intersect $B$ non-emptily: any point $x$ in $p \cap A$ is in $\overline{B}$ so any neighbourhood of $x$, including $p$, must intersect $B$.
$p\subset A\subset\bar A\subset\bar B$, then we must have $p\cap B\neq\emptyset$, then otherwise we would get $p\cap \bar B=\emptyset.$ Use the fact that $x\in\bar B$ $iff$ for each open set U containing $x$, $U\cap B\neq\emptyset$, and $p$ is open here.