$(X,\tau)$ needs to be $T_1$ in order to guarantee that $A'$ is closed? Claim: If $(X,\tau)$ is not $T_1$ then $A'$ is not necessarily closed for any $A$.
Proof: Here is an example. Take $X=\mathbb{N}$ and let $\tau=\lbrace u\subset\mathbb{N}\mid 1\in u\iff 2\in u\rbrace$. Let $A=\lbrace 2,3 \rbrace$. Since any open set containing $1$ also contains $2$ it follows that $1\in A'$. Hence $\overline{A}=\lbrace1,2,3\rbrace$, as it is the smallest closed set containing $A$. But  $X\setminus A'=\mathbb{N}\setminus\lbrace 1\rbrace$ is not open because it contains $2$ and not $1$. The set of accumulation points $A'$ is not closed.
So, in general, trying to prove that $A'$ is closed in a topological space is an effort in futility?
 A: I don't know if such spaces are interesting enough to have a name, so I will here call them DSC-spaces ("Derived Sets are Closed"):
Definition: A topological space $(X,\tau)$ is a DSC-space if (and only if) for all subsets $A\subset X$ the derived set $A' = \{x\in X : x \text{ is a limit point of } A\}$ is closed.
Then the DSC property lies strictly between $T_0$ and $T_1$, i.e. we have the
Proposition:


*

*Every $T_1$-space is a DSC-space.

*Every DSC-space is a $T_0$-space.

*There are DSC-spaces that are not $T_1$-spaces.

*There are $T_0$-spaces that are not DSC-spaces.


Proof: In a $T_1$-space, $x$ is a limit point of $A$ if and only if every neighbourhood of $x$ contains infinitely many points of $A$. For if $x$ has an open neighbourhood $U$ containing only finitely many points $\{a_1,\dotsc,a_n\}$ of $A$, then $V = U\setminus (\{a_1,\dotsc,a_n\}\setminus \{x\})$ is an open neighbourhood of $x$ that intersects $A$ at most in $x$, hence $x$ is not a limit point of $A$. Thus let $(X,\tau)$ be a $T_1$-space, $A\subset X$ and $x\in A''$. Then every open neighbourhood $U$ of $x$ contains a limit point $y$ of $A$. Since $U$ is open, it is a neighbourhood of $y$, and hence contains infinitely many points of $A$, hence $x$ is a limit point of $A$. Thus for $T_1$-spaces, we have $A'' \subset A'$, which means $A'$ is closed.
Let now $(X,\tau)$ be a DSC-space, and $x,y$ two distinct points of $X$. Then either $y$ has a neighbourhood $U$ not containing $x$, or $y\in \{x\}'$, and then $X\setminus \{x\}'$ is an open neighbourhood of $x$ not containing $y$. Thus every DSC-space is a $T_0$-space.
The Sierpiński space $X = \{0,1\}$ with the topology $\tau = \left\{\varnothing,\{1\},X\right\}$ is a $T_0$-space that is not a $T_1$-space, and it is a DSC-space. For $A = \varnothing$ or $A = \{0\}$, we have $A' = \varnothing$, and for $A = \{1\}$ or $A = X$ we have $A' = \{0\}$, so all derived sets are closed. Thus we see that the DSC property is strictly weaker than $T_1$.
Let $X = \mathbb{R}$, and
$$\tau = \{\varnothing\} \cup \left\{ M \subset \mathbb{R} : \bigl(\exists \varepsilon > 0\bigr)\bigl( (-\varepsilon,\varepsilon) \subset M\bigr)\right\}.$$
Then $\tau$ is a topology, and $(X,\tau)$ is a $T_0$-space: let $x,y$ be two distinct points in $X$; if $x\neq 0$ then $X\setminus \{x\}$ is an open neighbourhood of $y$ not containing $x$, otherwise $X\setminus \{y\}$ is an open neighbourhood of $x$ not containing $y$. But $(X,\tau)$ is not a DSC-space:
$$\{0\}' = X\setminus \{0\}$$
is not closed. Hence the DSC property is strictly stronger than $T_0$.
