My proof is correct ? (Topology) I want to prove that if for two open sets $A,B$ $A\cap B=\emptyset$ then $\overset{\circ}{\overline{A}}\cap \overset{\circ}{\overline{B}}=\emptyset$. 
I supposed that  $\overset{\circ}{\overline{A}}\cap \overset{\circ}{\overline{B}}\neq\emptyset$ then there exist $x\in   \overset{\circ}{\overline{B}} $ and $x\in   \overset{\circ}{\overline{A}}$. 
$x\in   \overset{\circ}{\overline{A}}$ means that $\overline{A}\in \mathcal{V}_x$ and 
$x\in   \overset{\circ}{\overline{B}}$ means that $x\in \overline{B}$ i.e., $\forall V\in \mathcal{V}_x,\ V\cap B\neq \emptyset$ in particular this is true for  $V=\overline{A}$.
So  $\overline{A}\cap B\neq \emptyset$. 
Then there exists $y\in \overline{A}$ and $y\in B$ as $B$ is open, $B$ is a ngbh of $y$, 
$y\in \overline{A}\Longleftrightarrow \forall W\in \mathcal{W}_y,\  W\cap A\neq \emptyset$  in particular this is true for $W=B$ so $$A\cap B\neq \emptyset.
$$ Is it correct ? The problem is that i don't use the fact that $A$ is open .
Thank you.
 A: The proof is as follow:
If $A,B$ are open and $A\cap B=\varnothing$, $A\cap\overline{B}=\varnothing$. If not, suppose $x\in A\cap\overline{B}$. Then there is an open set $G_x$ such that $x\in G_x$ and $G_x\subset A\cap\overline{B}$. So $G_x\subset\overline{B}$. Consider $\overline{B}=B\cup B'$, where $B'$ is the limit point set of $B$. So there is
$$
G_x\cap\overline{B}=G_x\cap (B\cup B')=(G_x\cap B)\cup (G_x\cap  B')\ne \varnothing
$$
This means that either $G_x\cap B\ne \varnothing$ or $G_x\cap B'\ne \varnothing$. But $G_x\cap B'\ne \varnothing$ means $G_x\cap B\ne \varnothing$ for there is at least a point of $B$ in $G_x\cap B'$. But this contradicts $A\cap B=\varnothing$. So we have $A\cap\overline{B}=\varnothing$. Since $A\cap\overset{\circ}{\overline{B}}\subset A\cap\overline{B}=\varnothing$, we have $A\cap\overset{\circ}{\overline{B}}=\varnothing$.
Likewise, we can prove by the same reasoning that $\overline{A}\cap\overset{\circ}{\overline{B}}=\varnothing$ ($\overset{\circ}{\overline{B}}$ is open). Since $\overset{\circ}{\overline{A}}\cap\overset{\circ}{\overline{B}}\subset\overline{A}\cap\overset{\circ}{\overline{B}}=\varnothing$, finally we have $\overset{\circ}{\overline{A}}\cap\overset{\circ}{\overline{B}}=\varnothing$.
A: I cannot follow your proof. You could start by noticing that if $A$ and $B$ are disjoint and if $A$ is open then $A\cap \bar B=\phi.$ Because any $a\in A$ has a nbhd (namely,$A$) which is disjoint from $B$.  Now if $x\in \bar A$ and if $V $ is any nbhd of $x$, then $V\cap A\ne \phi$ and $(V\cap A)\cap \bar B\subset A\cap \bar B=\phi $. So no nbhd $V$ of $x$ is a subset of $\bar B$.That is,for all $x$ we have $$x \in \bar A\implies x\not \in int(\bar B).$$ Therefore $$int(\bar A)\cap int(\bar B)\subset \bar A\cap int(\bar B)=\phi.$$ You must use the fact that at least one of $A,B$ is open.I did not require that $B$ be open.   
