Well, take the two-point indiscrete space $X=\{a,b\}$ and let $A=\{a\}$. Then $A'=\{b\}$, which is not closed.
Here is a more sophisticated example which is even $T_0:$
Let $X=[-1,1]$ and for $0$ take as a neighborhood base
$$\mathcal N_0=\{(-\varepsilon,\varepsilon)\mid \varepsilon>0\}$$
for an $x\ne0$ take as a local base
$$\mathcal N_x=\{\{x\}\cup(-\varepsilon,\varepsilon)\mid \varepsilon>0\}$$
It is easy to verify the axioms for a system of neighborhood filter bases.
Now, $A=\{0\}$ has the derived set $X-\{0\}$ which is not closed.
Here is a proof that in a finite $T_0$ space, and more general in a $T_0$ space where each point has a smallest neighborhood, all derived sets are closed:
Let $X$ be such a space, and $A\subset X.$ Let $y\in\overline{A'}$. If $U$ is the smallest neighborhood of $y,$ then $U$ is open and contains some $x\in A'.$ By $T_0$ there is a neighborhood $V$ of $x$ such that $y\notin V$. Since $U$ is open, $U\cap V$ is also a neighborhood of $x$ and must intersect $A-\{y\}.$ This shows that each neighborhood of $y$ intersects $A-\{y\},$ hence $y\in A'.$
Your example isn't quite right: $\{\{1\},\{1,2,3\}\}$ is not in fact a topology. You need to throw in the emptyset: $\{\emptyset, \{1\},\{1,2,3\}\}$ is a topology.
The definition on wikipedia is correct - the issue is what exactly is meant when we say "$\tau$ is a topology on $X$ in the sense of Hausdorff."
I suspect that you are getting tripped up by interpreting this as "$\tau=ran({\bf N})$ for some ${\bf N}$ satisfying Hausdorff's axioms." But that's not correct. Instead, we need to introduce yet another notion:
$U\subseteq X$ is open according to ${\bf N}$ iff for each $x\in U$ we have some $V\in {\bf N}(x)$ such that $V\subseteq U$.
(On the wikipedia page, this is presented in the last paragraph of the "definition by neighborhoods" section.)
This notion is what provides the bridge between the "neighborhood function" idea of topology and the "set of open sets" idea of topology: when we say
"$\tau$ is a topology on $X$ in the sense of Hausdorff,"
what we mean is
"there is some ${\bf N}$ satisfying Hausdorff's axioms such that $\tau$ is the set of open sets in the sense of ${\bf N}$."
Note that if $U$ is open in the sense of ${\bf N}$ and $x\in U$ then $U\in {\bf N}(x)$, but the converse is false: you want to interpret "neighborhood of $x$" as "set containing some open set containing $x$." It is true (as you observe) that $\{\emptyset,\{1\},\{1,2,3\}\}$ is not $ran({\bf N})$ for any ${\bf N}$ satisfying Hausdorff's axioms, but that's irrelevant.
In particular, to show that $$\{\emptyset, \{1\},\{1,2,3\}\}$$ is a topology on $\{1,2,3\}$ in the sense of Hausdorff, we need to exhibit an $${\bf N}:\{1,2,3\}\rightarrow\mathcal{P}(\{1,2,3\})$$ satisfying Hausdorff's neighborhood function axioms such that the ${\bf N}$-open sets are exactly $\emptyset$, $\{1\}$, and $\{1,2,3\}$. This might look daunting, but luckily there's a general recipe for doing this. Given any collection $\mathcal{S}$ of subsets of $X$, consider the function $${\bf N}_\mathcal{S}: x\mapsto\{U\in\mathcal{S}: x\in U\}.$$ It turns out (and this is a good exercise) that this is the only thing we need to consider: $\mathcal{S}$ is a topology on $X$ in the usual sense iff ${\bf N}_\mathcal{S}$ satisfies Hausdorff's neighborhood function axioms (in which case $\mathcal{S}$ will be exactly the collection of open sets in the sense of ${\bf N}_\mathcal{S}$).
In this case we're led to consider the function ${\bf N}: \{1,2,3\}\rightarrow\mathcal{P}(\{1,2,3\})$ given by:
$1\mapsto \{\{1\},\{1,2\}, \{1,3\}, \{1,2,3\}\}$,
$2\mapsto\{\{1,2,3\}\}$,
$3\mapsto \{\{1,2,3\}\}$.
It's easy to check that this ${\bf N}$ satisfies Hausdorff's axioms and that the open sets in the sense of ${\bf N}$ are exactly $\emptyset$, $\{1\}$ and $\{1,2,3\}$
Best Answer
If $X$ is a $T_1$ space then $S'$ is closed.
Suppose by contradiction that $p\in \overline {S'}$ \ $S'.$ Since $S'\subset \bar S$ we have $p\in \bar S.$
Now $p\not \in S'$ implies there is an open $U$ with $p\in U$ and $U\cap S\subset \{p\}.$
But $U$ must contain $q\in S'$ because $p\in \overline {S'}$, and by hypothesis, $p\ne q.$ Then there is an open $V$ with $q\in V$, and $p \not \in V$, because $X$ is $T_1.$ The open set $U\cap V$ contains $q\in S'$, so there exists $r\in U\cap V\cap S$ with $r\ne q.$ BUT then $p\ne r\in S\cap U,$ contrary to $U\cap S \subset \{p\}$.
As Opn Ball has already pointed out, the property does not hold for the coarse topology on $X$ if $X$ has at least 2 points: For $S=\{p\}\subset X$ we have $S'=X$ \ $\{p\}$ and $\overline {S'}=X\ne S'.$