Does it hold that, if every set $A$ in a topological space $X$ satisfies $A'=\bar{A}'$, then $X$ is a T0-topological space?
The equality doesn't imply T1: Sierpiński space $S$ is T0 but not T1 since $A=\{1\}$ is not closed.
$A$ is the only non-closed set, and we have $\bar{A}=S$ and $\bar{A}'=\{0\}=A'$.
