closure of open set in topological space How to prove : If $X$ is a topological space, $U$ is open in $X$, and $A$ is dense in $X$, then $\overline{U}=\overline{U \cap A}$
 A: Note that 
$$
\overline U = \underset{ \mathcal F \text{ closed}}{ \bigcap_{U \subseteq \mathcal F} }\mathcal F, \quad \overline {U \cap A}  = \underset{ \mathcal F \text{ closed}}{ \bigcap_{U \cap A \subseteq \mathcal F} }\mathcal F, 
$$
so to see that $(\subseteq)$ and $(\supseteq)$ holds, it suffices to see that $U \subseteq \mathcal F$ if and only if $U \cap A \subseteq \mathcal F$ when $\mathcal F$ is closed, $U$ is open and $A$ is dense in $X$. One direction is clear because $U \cap A \subseteq U$. For the other one, note that if $U \cap A \subseteq \mathcal F$, then $A \subseteq \mathcal F \cup U^c$, and since $U$ is open, $\mathcal F \cup U^c$ is closed, hence $X = \overline A \subseteq \mathcal F \cup U^c$ and $\mathcal F \cup U^c = X$, which means $U \cap \mathcal F^c = \varnothing$, thus $U \subseteq \mathcal F$. 
Hope that helps,
A: A slightly different demonstration (though easily translatable into/from Patrick Da Silva's answer) of this fact is as follows:

First note that as $U \cap A \subseteq U$ then $\overline{U \cap A} \subseteq \overline{U}$, so we need only show the reverse inclusion.  Take any $x \in \overline{U}$.  To show that $x \in \overline{U \cap A}$ we need only show that $V \cap ( U \cap A ) \neq \emptyset$ for every open neighbourhood $V$ of $x$.  Given an open neighbourhood $V$ of $x$, since $x \in \overline{U}$, then $V \cap U \neq \emptyset$. But $V \cap U$ is an open set, now known to be nonempty, and since $A$ is dense we must have $V \cap ( U \cap A ) = ( V \cap U ) \cap A \neq \emptyset$.  It thus follows that $x \in \overline{ U \cap A }$, and so $\overline{U} \subseteq \overline{U \cap A}$.

