TITLE: If A and B are disjoint open sets, prove that they are separated
QUESTION [6 upvotes]: I just proved this statement like the below.
Is this valid or solid proof?
Thank you!

REPLY [8 votes]: Your idea looks fine. The proof that $\overline A\cap B=\varnothing$ can be direct, and by symmetry (as you say), we also get $A\cap\overline B=\varnothing$.

Take $b\in B$. Since $B$ is open, there is some open neighborhood $N$ with $b\in N\subseteq B$. But then from $A\cap B=\varnothing$ we get $N\cap A=\varnothing$; so $b\notin\overline A$.
(Yes, we can even take $B=N$!)

ADD Recall that $x\in \overline A$ if and only if for each open set $O$ contaning $x$, we have $A\cap O\neq \varnothing$.
In fact, we can say more:

Let $(X,\mathscr T)$ be a (topological, metric) space. Then the following definitions of connectedness are equivalent:
$(1)$ There exist two non-empty disjoint open sets $A,B$ with $A\cup B=X$.
$(2)$ There exist two non-empty disjoint closed sets $A',B'$ with $A'\cup B'=X$.
$(3)$ There exist two non-empty separated sets $A'',B''$ with $A''\cup B''=X$.