Let
$$X=\{a_{ij},b_{ij},c_i,a,b\mid i=1,2,\ldots;\,j=1,2,\ldots\}$$
where all these elements are assumed to be distinct. Define the following neighborhood systems on $X$ : Each $a_{ij}$ is isolated and each $b_{ij}$ is isolated:
$$\mathscr{B}(c_i)=\left\{V^n(c_i)=\bigcup_{j=n}^\infty\{a_{ij},b_{ij},c_i\}\mid n=1,2,\ldots\right\}$$
$$\mathscr{B}(a)=\left\{V^n(a)=\bigcup_{j=1}^\infty\bigcup_{i=n}^\infty\{a_{ij},a\}\mid n=1,2,\ldots\right\}$$
$$\mathscr{B}(b)=\left\{V^n(b)=\bigcup_{j=1}^\infty\bigcup_{i=n}^\infty\{b_{ij},b\}\mid n=1,2,\ldots\right\}\,.$$
Let us denote this topology by $\tau$.
Why is $(X,\tau)$ is minimal Hausdorff but not compact?
$X$ is clearly not compact: $\{V^1(a),V^1(b)\}\cup\{V^1(c_i):i\in\Bbb Z^+\}$ is an open cover with no finite subcover.
A space is minimal if and only if it is semiregular and H-closed, as is noted in the remarks at the end of the answer to this question. It is straightforward to verify that $X$ is H-closed. If $\mathscr{U}$ is an open cover of $X$, there are $U_a,U_b\in\mathscr{U}$ such that $a\in U_a$ and $b\in U_b$. From the definition of $\tau$ it’s clear that there is a $k\in\Bbb Z^+$ such that
$$a_{ij},b_{ij},c_i\in\operatorname{cl}U_a\cup\operatorname{cl}U_b$$
for all $i>k$ and all $j\in\Bbb Z^+$. For $i=1,\ldots k$ there are a $U_i\in\mathscr{U}$ and an $n_i\in\Bbb Z^+$ such that $c_i,a_{ij},b_{ij}\in U_i=\operatorname{cl}U_i$ for all $j\ge n_i$. Thus,
$$X\setminus\left(\operatorname{cl}U_a\cup\operatorname{cl}U_b\cup\bigcup_{i=1}^k\operatorname{cl}U_i\right)$$
is at most a finite set of isolated points and is covered by the closures of some finite $\mathscr{F}\subseteq\mathscr{U}$, so that
$$X=\operatorname{cl}U_a\cup\operatorname{cl}U_b\cup\bigcup_{i=1}^k\operatorname{cl}U_i\cup\bigcup\{\operatorname{cl}U:U\in\mathscr{F}\}\,,$$
and $X$ is therefore H-closed.
Finally, all of the basic open nbhds $\{a_{ij}\}$, $\{b_{ij}\}$, and $V^n(c_i)$ are clopen, and the basic open nbhds $V^n(a)$ and $V^n(b)$ are regular open, so $X$ is semiregular.