By $A'$ I mean the set of limit points of $A$, so $A''$ is the set of limit points of $A'$.
There is no T1 space with such a subset. Proof:
Suppose $X$ is a T1 space and let $A \subset X$. Let $x \in A''$ and $U$ be an open neighborhood of $x$. By definition, there exists $y \in U \cap (A' \setminus \{x\})$. The set $U \setminus \{x\}$ is an open neighborhood of $y$ because $X$ is T1, hence there is a point in $U \setminus \{x\} \cap A - \{y\}$. This is also a point in $U \cap A - \{x\}$, so $x \in A'$.
There is a non-T0 space with such a subset: let $X = \{a,b\}$ have the indiscrete topology, and pick $A = \{b\}$. Then $A' = \{a\}$ and $A'' = \{b\}$.
My question is if there is a T0 (but not T1) space with a subset with this property. It does not seem to be possible with the standard T0 spaces I have checked (particular/excluded point spaces, and other simple spaces).