Closure and inclusions Let $A$ and $B$ be subsets of $\mathbb{R}^n$. If $A$ is open, and $B$ is arbitrary, does one always have the following inclusion $A \cap \operatorname{cl}(B) \subseteq \operatorname{cl}(A\cap B)$.

Is there a counter-example? I can't seem to find a way to prove it.
The statement is true. The proof just uses the relevant definitions. 

Suppose $x \in A \cap \operatorname{cl} (B)$, i.e., $x \in A$ and $x \in \operatorname{cl}(B)$. The latter condition means that if $U$ is an open set containing $x$, then $U$ intersects $B$ (nontrivially). 

Now, pick any open set $V$ containing $x$. Then $V \cap A$ is an open set containing $x$. Then by the preceding paragraph, $(V \cap A) \cap B \neq \emptyset$, which is equivalent to $V \cap (A \cap B) \neq \emptyset$. This means that $x \in \operatorname{cl}(A \cap B)$.
the closure is closed, and the intersection of a closed set and an open set is not always a closed set

a conter exemple may be: A=]0,2[ B=]1,3[