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.
Munkres' proof certainly uses AC. The alternative proof doesn't explicitly use AC, but as Asaf Karagila remarks in his comment, it may be hidden in the proof that compact Hausdorff spaces are normal.
Frankly speaking, I believe that most of us are adherents of ZFC, and I personally did not spend much time in questions concerning the use of AC. However, in this case I tried to find a proof without using AC. So let $X$ be a compact Hausdorff space.
1) $X$ is regular.
Let $x \in X$ and $B \subset X$ be closed such that $x \notin B$. For $y \ne x$ let us say that an open neighborhood $U$ of $y$ is of type $H$ (for Hausdorff) if there exists an open neighborhood $V$ of $x$ such that $U \cap V = \emptyset$. Clearly, each $y \ne x$ has such a neighborhood. Let $\mathfrak{U}(y)$ be the set of all open neighborhoods $U$ of $y$ of type $H$ and $\mathfrak{U} = \bigcup_{y \in B} \mathfrak{U}(y)$. This is a cover of $B$ by open sets in $X$. Since $B$ is closed in $X$, it is compact and there exist finitely many $U_i$ in $\mathfrak{U}$ such that $B \subset U^* = \bigcup_{i=1}^n U_i$. Now we can make finitely many choices to get open neigborhoods $V_i$ of $x$ such that $U_i \cap V_i = \emptyset$. Then $V^* = \bigcap_{i=1}^n V_i$ is a an open neighborhood of $x$ such that $U^* \cap V^* = \emptyset$.
As far as I can see this does not use AC. The "standard proof", however, is based on AC by choosing for each $y \ne x$ a pair of open neigborhoods $U_{y}$ of $y$ and $V_{y}$ of $x$ such that $U_{y} \cap V_{y} = \emptyset$.
2) $X$ is normal.
Let $A, B \subset X$ be closed such that $A \cap B = \emptyset$. For $y \notin B$ let us say that an open neighborhood $U$ of $y$ is of type $R$ (for regular) if there exists an open neighborhood $V$ of $B$ such that $U \cap V = \emptyset$. By 1) each $y \notin B$ has such a neighborhood. Adapting the proof of 1), we see that that $A, B$ have disjoint open neighborhoods.
I hope I didn't make a mistake in showing without AC that "compact Hausdorff $\Rightarrow$ normal". But in my opinion the definition of compactness resembles the spirit of AC. It allows to make a choice: For each open cover it assures the existence of a finite subcover, but it is entirely nebulous how this finite subcover can be found. Of course, all finite $X$ are compact, but to prove the compactness of an infinite space $X$ in many cases AC is needed. For example the compactness of infinite products of compact spaces is equivalent to AC.
Best Answer
This is from "General topology" by Riszard Engelking.
It seems the following.
Let $A$ be a minimal element of the family $\mathcal{A}$. Suppose that the set $A$ is not connected. Then $A$ is a union of two its disjoint clopen non-empty subsets . Since $x$ and $y$ are in the same quasicomponent of $A$, the set $\{x,y\}$ is contained in one of these two sets. Denote this set of $B$. Since the set $B$ is a clopen subset of $A$, each quasicomponent of the set $B$ is a quasicomponent of the set $A$. Hence $B\in\mathcal A$. But $B$ is a proper subset of the set $A$, which contradicts to the minimality of the set $A$.