Closed Intersection of sets is contained in closed sets intersection $\overline{A{\cap}B}$ $\subseteq$ $\overline{A}\cap\overline{B}$
What I have so far is let x$\in$ $\overline{A{\cap}B}$, then x$\in$A$\cap$B and x$\in(A$$\cap$$B)'$. But Im not sure where to go from there...
 A: Let $x\in \overline{A\cap B}$.  Then every neighborhood of $x$ meets $A\cap B$, therefore every neighbhorhood of $x$ meets $A$. We see that $x\in \overline{A}$; therefore, $\overline{A\cap B} \subseteq \overline{A}$.  A symmetric argument shows $\overline{A\cap B} \subseteq \overline{B}$
A: Let $x\in\overline{A\cap B}$. For every $\epsilon>0$ there is $y\in B(x,\epsilon)\cap A\cap B$. Therefore $y\in B(x,\epsilon)\cap A$ and $y\in B(x,\epsilon)\cap B$.   So $x\in\overline A\cap\overline B$.
A: Let $x$ be in $\overline{A \cap B}$. So there is a sequence $\{x_k\} \subset A \cap B$ such that $x_k \rightarrow x$. Besides, $\{x_k\} \subset A$ results that $x \in \overline{A}$, and $\{x_k\} \subset B$ results that $x \in \overline{B}$. Therefore, $x \in \overline{A} \cap \overline{B}$.
A: Use the property that the closure of a set is the union of all its closed supersets:
$A\cap B\subset A \subset \bar A \implies \overline{A\cap B}\subset\bar A$ analogously $\overline{A\cap B}\subset\bar B$, so $\overline{A\cap B}\subset\bar A \cap \bar B$
