I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $v\in\{2,3,4\}$.
Here is a translation of H. Herrlich's proof:$\newcommand{\mc}[1]{\mathcal{#1}}$
Let $(X,\mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',\mc T')$ such that $X'=X\cup\{a\}$ and $X$ is not closed in $(X',\mc T')$. If we choose an arbitrary element $x_0\in X$ then
$$\mc T''=\{M; M\in\mc T; x_0\in M \Rightarrow M\cup\{a\}\in\mc T'\}$$
is a $T_2$-topology on $X$ which is strictly weaker than $\mc T$. Hence $(X,\mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $\mc T''$ is Hausdorff: If we have $x_0\ne y$, $y\in X$ then there are $\mc T'$-neighborhoods $U_x\ni x$, $V_1\ni y$ which are disjoint. Similarly, we have $U_a\ni a$, $V_2\ni y$, which are disjoint. Hence $U_x\cup (U_a\cap X)$ and $V_1\cap V_2$ are $\mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $\mc T$.
The fact that $\mc T''$ is strictly weaker than $\mc T$ follows from the fact, that $\{a\}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,\mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}\ni x$ and $U_a\ni a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,\mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $\newcommand{\N}{\mathbb N}\N^*=\{0\}\cup\{\frac1n; n\in\mathbb N\}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $\N\times\N^*$, where $\N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)=\{(n,1/m)\in\N\times\N^*; n\ge n_0\}\cup\{q\}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=\operatorname{Int} \overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $n\ge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
Any topology with finitely many open sets must have the property that all sets are compact, simply because any open cover is already finite.
Here is a much less trivial example:
Let $(\mathbb R,\tau)$ be the real numbers with the co-finite topology, namely a set is closed if and only if it is finite; and a set is open if and only if its complement is finite.
Consider the natural numbers as a subset of the real line, this is an infinite set, but it is clearly not co-finite so it is neither open nor closed. Suppose that $\{U_i\mid i\in I\}$ is an open cover of $\mathbb N$. There is some $i_0\in I$ such that $0\in U_{i_0}$, and since $U_{i_0}$ is open it means that it contains everything except finitely many points, in particular it must contain all the natural numbers, except maybe finitely many of them. For every $n\in\mathbb N\setminus U_i$ we can find some $U_{i_n}$. We found, therefore, a finite subcover of this open cover, and so $\mathbb N$ is compact.
Exercise: Prove that in fact every set of real numbers is compact in this topology.
Best Answer
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 $y\in A'.$