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.
It is possible to deduce ($2$) from ($3$) without $q\times q$ being a quotient map.
Let $A\subset X$ be closed. Assume $R=\{(x,x')\in X\times X\mid q(x)=q(x')\}$ is closed, thus compact. Then $A\times X\cap R$ is compact, and so is also its image under the projection $p_2$ onto the second factor. But $p_2(A\times X\cap R)=\{x\in X\mid\exists a\in A:q(a)\sim q(x)\}=q^{-1}(q(A))$. So the saturation of a closed set is compact, and hence closed, which means that $q$ is a closed map.
One could also omit the compactness of $X$ if one assumes directly that $R$ is compact, because the last step only uses the Hausdorff'ness to deduce that a compact set is closed. Note that compactness of $R$ also makes $\{x\}\times X\cap R$ a compact set, so fibers are compact and this makes $q$ a so-called perfect map. These maps preserve many properties of the domain, for example all the separation axioms (except $T_0$)
This doesn't answer the question in the title, which I would really like to know myself. I only know about the product of a quotient map with the Identity $q\times Id:X\times Z\to Y\times Z$, which is a quotient map if $Z$ is locally compact.
Edit: Actually, showing that $q$ is closed requires only the compactness of $X$. Indeed, if $X$ is compact, then the projection $p_2$ is a closed map, so if $R$ is closed, then for every closed $A\subseteq X$, the set $p_2(A\times X\cap R)$ is closed.
Best Answer
Here's a counter-example.
Let $X = \hat{\mathbb{Q}}$ be the one-point compactification of $\mathbb{Q}$. It's well-known that such compactification is Hausdorff iff the underlying space is locally compact and Hausdorff, so $X$ is not Hausdorff.
If $Y\subseteq X$ is closed then it's compact as a closed subset of compact space, so we only need to consider if compact subsets are closed.
Let $Y\subseteq X$ be compact. If $Y\subseteq \mathbb{Q}$ then $Y$ is closed in $X$ by definition. Otherwise we can write $Y= Y_0\cup \{\infty\}$ where $Y_0\subseteq \mathbb{Q}$. It suffices to prove that $Y_0$ is closed in $\mathbb{Q}$. From compactness of $Y$ it follows that if $K\subseteq \mathbb{Q}$ is compact then $Y_0\cap K$ is compact. To see this, cover it by any family of open subsets of $\mathbb{Q}$ and add $(\mathbb{Q}\setminus K)\cup \{\infty\}$ to it. From compactness of $Y_0$ we can extract an open subcover, and we can remove the additional set as it doesn't intersect $Y_0\cap K$. Take a sequence $y_n\in Y_0$ convergent in $\mathbb{Q}$ to $y$, and consider $K = \{y_n: n\in\mathbb{N}\}\cup \{y\}$, then from compactness of $Y_0\cap K$ we have $y\in Y_0$. This proves that $Y_0$ is closed in $\mathbb{Q}$.
Note that I haven't used any properties other than that $\mathbb{Q}$ is a metric space which is not locally compact. This gives us a whole family of counter-examples.