if $A$ and $B$ are open and $A \cap B = \emptyset$, then $A \cap \overline{B} = \overline{A} \cap B = \emptyset$ if $A$ and $B$ are open and $A \cap B = \emptyset$, then  $A \cap \overline{B} =  \overline{A} \cap B = \emptyset$
Assume $A$ and $B$ are open and $A \cap B = \emptyset$. Suppose $\overline{A} \cap B \not= \emptyset$. Then there exists some $x$ such that $x \in \overline{A}$ and $x \in B$. If $x \in \overline{A}$ then any epsilon-ball $B_\epsilon (x)$ will contain points in $A$. Since B is open, $B_\epsilon (x) \subseteq B$. This means that there are points in that are in both $A$ and $B$, which contradicts the assumption that $A \cap B = \emptyset$. Then you can do the same then for the case where $A \cap \overline{B} \not= \emptyset$
Would this be an ok proof of this? If not, what can I improve?
 A: Based on your comments, you're dealing with $\mathbb{R}^n$, but you really need to make sure you're defining the space you're working with, either in the theorem or the proof. Your proof is generally correct, but it's sloppy. You should be saying, "Since $B$ is open, there exists $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq B$, and this will contain points in $A$." Also, this is mostly a style thing, but I would recommend saying at the end that the argument for $A\cap \overline{B}\neq 0$ is identical.
Since it seems you're working with analysis, I should point out that this is true for any metric space, not just $(\mathbb{R}^n,d)$, where $d$ is the standard metric (distance formula), and your proof highlights this clearly. I would recommend looking into metric spaces, when you get a chance; this will make a lot more sense then.
A: Your proof is nearly correct and has right the right idea. The one problem is when you use the arbitrary epsilon ball as the ball that will be contained in B. It would be better to start like this: "Suppose x $\in \bar{A}$ and $x \in B$. Then since $B$ is open, there is an $\epsilon > 0$ such that $B_{\epsilon}(x) \subset B$..." Now use the fact that $x$ is in the closure of $A$ to proceed to the contradiction.
A: Getting down into the details, like you did in your proof, is a great way to get used to the concepts, but I will point out that there's an almost trivial proof if you use some basic properties of sets and topology:
Since $A \cap B = \emptyset$, we know that $B \subset A^C$ (where $A ^ C$ is the complement of $A$).  Now there's a general rule that if $D$ and $E$ are arbitrary sets then
$$ D \subset E \implies \overline D \subset \overline E,$$
so 
$$ \overline B \subset \overline{A^C} = A^C,$$
because $A$ being open means that $A^C$ is closed. But this implies that $A \cap \overline B = \emptyset$.
A: In any space $X,$ if $B\subset X$ and $a\in X$ we have: 
$a\not \in \overline B$ iff there exists an open set $U_a \subset X$ which contains $a$ and is disjoint from $B.$
Now if $A$ is open and disjoint from $B$ then for any $a\in A$ we can take $U_a=A,$ and conclude that no $a\in A$ belongs to $\overline B.$
Similarly with $A,B$ interchanged.
