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.
Best Answer
Here is a nice example. Let $X$ be a set with $|X|>2$, and give $X$ the trivial topology (so the only open sets are $\varnothing$ and $X$). Let $a,b\in X$ be distinct points. Then the map $f:X\to X$ with $f(a)=b$, $f(b)=a$, and $f(x)=x$ for any $x\neq a,b$ has as its set of fixed points $X\setminus\{a,b\}$, which is not closed because the only closed sets are $\varnothing$ and $X$.
It is easy to see that $X$ is not Hausdorff; because the only open sets available to us are $\varnothing$ and $X$, we can't separate distinct points with disjoint open sets.