Does the derived set of the closure of a set $A$ equal the derived set of $A$? I have been given this task for homework and I simply cannot wrap my head around it. 

Does it hold that  $\mathrm{ cl}(A)'=A'$, where $A'$ denotes the derived set of $A$ and $\mathrm{cl}$ is the closure operator? 

At first I tried to find some counterexamples, but since I could not come up with any, I assumed it must be true. Now I am struggling with proving it. 
 A: For simplicity, I shall write $\bar{A}$ for the closure of a set $A$.  As a counterexample, let $X:=\{1,2\}$ be the set equipped with the trivial topology $\mathcal{T}:=\{\emptyset,X\}$.  Then, with the subset $A:=\{1\}$ of $X$, we have $\bar{A}=X$ and $\bar{A}'=X$.  However, $A'=\{2\}$.
The direction $A'\subseteq \bar{A}'$ is true and easy to prove.  If $x\in A'$, then for every open set $U$ with $x\in U$, $U$ intersects $A$ in a point $y\neq x$.  Since $A\subseteq \bar{A}$, $y\in \bar{A}$.  This finishes the proof of this direction.
The other direction ($\bar{A}'\subseteq A'$) is not true, as noted above.  Nonetheless, if $X$ is a $T_1$-space (i.e., singletons are closed sets), then the claim holds.  To show this, let $x\in \bar{A}'$ be arbitrary.  We prove via contradiction by assuming on the contrary that there exists an open set $U$ with $x\in U$ such that $U\setminus\{x\}$ is disjoint from $A$.  Thus, $U\setminus \{x\}$ is an open set (this is where the $T_1$ assumption is used) disjoint from $A$.  Therefore, by the definition of topological closures, $U\setminus \{x\}$ must be disjoint from $\bar{A}$ as well.  However, by the definition of limit points and the choice of $x$, $U$ must contain a point in $\bar{A}$ other than $x$.  This is a contradiction.


Interesting Question: Does it hold that, if every set $A$ in a topological space $X$ satisfies $A'=\bar{A}'$, then $X$ is a $T_1$-topological space?

It turns out that the answer to the question above is also no.  For those who are interested, see the hidden portion below.

  Let $X$ be the set $\{1,2\}$, as before, but now equipped with the topology $\tilde{\mathcal{T}}:=\big\{\emptyset,\{1\},X\big\}$.  The only subset $A$ of $X$ which is not closed is $A:=\{1\}$, and we have $\bar{A}=X$ and $\bar{A}'=\{2\}=A'$.

