Let $T =(X, \tau)$ an extremally disconnected topological space, i.e. for every $A \in \tau$, $\overline{A} \in \tau $. Thus, $T$ is also a $T_1$ space? If not, there exists a simple counterexample?
In this definition, $T$ can be a non-Hausdorff space.
The simplest non-$T_1$ space is already an example: the indiscrete topology on any space, as the only open subsets $\emptyset$ and $X$ are also closed.
The Sierpiński topology on $\{0,1\}$ also works: the only open sets are $\{0\}$ and $\{0,1\}$ and $\emptyset$ with respective closures $X=\{1,2\}$,$X$ and $\emptyset$.
Usually extremally disconnected spaces are assumed to be Hausdorff. In that case it implies total disconnectedness etc. The examples I have are connected extremally disconnected spaces. This is already linguistically odd and illogical at the least. Of course we can take a disjoint sum of many non-trivial indiscrete spaces to get a disconnected extremally disconnected space that is not $T_0$ nor $T_1$.