TITLE: A topological space is extremally disconnected iff every two disjoint open sets have disjoint closures
QUESTION [8 upvotes]: Show that for any topological space $X$ the following are equivalent:

$X$ is extremally disconnected
Every two disjoint open sets in $X$ have disjoint closures.


My attempt at a solution
Assume every two disjoint open sets in $X$ have disjoint closures.
Let A open in $X$, $A$ and $X \setminus \overline{A} $  disjoint open set
$\overline{A} \cap (\overline{X \setminus \overline{A}} )= \emptyset$,
$(\overline{X \setminus \overline{A}} ) \subset X \setminus \overline{A}$,
which implies that $X \setminus \overline{A}$ is closed in $X$. Thus $\overline{A}$ is open in $X$. So we obtain $X$ is extremally disconnected.
Otherwise Let $X$ is extremally disconnected. Then the closure of every open set is open.
 So $X$ is completely separated. Let $A$, $B$ two disjoint open sets.  Then there exist a continuous function $g:X \rightarrow [0,1]$ such that  $g(A)=0$, $g(B)=1$
How can I continue?

REPLY [4 votes]: The first half of your argument is fine. For the second half you don’t need the function $g$. You have the disjoint open sets $A$ and $B$. If $x\in B$, then $B$ is an open nbhd of $x$ disjoint from $A$, so $x\notin\operatorname{cl}A$. Thus, $B\cap\operatorname{cl}A=\varnothing$. Now suppose that $x\in\operatorname{cl}A$; $\operatorname{cl}A$ is open, so it’s an open nbhd of $x$ disjoint from $B$, and $x\notin\operatorname{cl}B$. It now takes just one small step to finish the argument to show that $(\operatorname{cl}A)\cap\operatorname{cl}B=\varnothing$.