Prove a set $A\subset X$ is open iff for every $B\subset X$: $\overline{A\cap \overline{B}}=\overline{A\cap B}$. 
Prove a set $A\subset X$ is open iff for every $B\subset X$, $\overline{A\cap \overline{B}}=\overline{A\cap B}$.

I tried this as I would have tried to prove any two sets are equal but I can't find the way to the end. 
This is homework.
 A: First note that for any set $B$, $A\cap B\subset A\cap\overline B$, so $\overline{A\cap B}\subset \overline{A\cap\overline B}$. So only the opposite inclusion matters.
Let's start with sufficient condition : suppose $\overline{A\cap\overline B} = \overline{A\cap B}$ for any set $B\subset X$. Then for $B=\complement A$, $\overline{A\cap\overline{\complement A}} = \overline{A\cap \complement A}=\emptyset$. This means $A\cap\overline{\complement A}=\emptyset$.
So $\complement(\complement A)=A\subset \complement(\overline{\complement A})$, and taking complement : $\overline{\complement A}\subset \complement A\subset \complement A$, so $\complement A$ is closed, meaning $A$ is open.
Now for the necessary condition : suppose $A$ open, and let $B$ be any set. We have to prove that $\overline{A\cap\overline B}\subset \overline{A\cap B}$. Let $x$ be a point of $\overline{A\cap\overline B}$. So any open set $V$ containing $x$ must verify $V\cap(A\cap\overline B)\ne\emptyset$.
But this means $(V\cap A)\cap\overline B\ne\emptyset$. As $V\cap A$ is an open set, it means $V\cap A$ meets $B$, so $V\cap(A\cap B)\ne\emptyset$. And that, finally, means $x\in\overline{A\cap B}$.
I hope I didn't say too many sillyness, my general topology is a bit rusty :-)
A: Suppose $A$ is open. 
It is clear that we must always have $\overline{A \cap \overline{B}} \supset \overline{A \cap B}$.
To prove the reverse inclusion, suppose $x \in \overline{A \cap \overline{B}} $. Given an open set $U \ni x$, there exists some $y \in A \cap \overline{B} \cap U$. Since $V = A \cap U$ is an open set about $y$ and $y \in \overline{B}$ there exists a $z \in B \cap V = A \cap B \cap U$. Thus, $A \cap B$ meets $U$, so by the arbitrary nature of the open set $U \ni x$, we find that $x \in \overline{A \cap B}$. This proves that $\overline{A \cap \overline{B}}  \subset \overline{A\cap B}$
We deduce that, under the hypothesis that $A$ is open, we have equality $\overline{A \cap \overline{B}}  = \overline{A\cap B}$. What's left is to prove that, if said equality is true, then $A$ must be open.
A: For $B=A^c$, we have $\overline{A \cap \overline{A^c}} = \overline{A\cap A^c} = \phi$ and hence $A \cap \overline{A^c} = \phi$ and $A \subset \overline{A^c}^c$. Also $A^c \subset \overline{A^c}$ and hence $\overline{A^c}^c \subset A$. Thus $A = \overline{A^c}^c $ and $A$ is open. The other part has been proved in the solution by joeb.
