# star compact Hausdorff is countably compact

I have read in some papers stated that every star compact Hausdorff is countably compact. I don't know how to prove it.

Note: A space $X$ is called star compact if for any open cover $\mathcal{U}$ of $X$, there exists finite subset $A\subseteq X$ such that $st(A,\mathcal{U})=\bigcup\{U\in\mathcal{U}:U\cap A\neq\emptyset\}=X$

This is theorem 2.1.5 in this paper, Star covering properties, by E.K. van Douwen, G.M. Reed, A.W. Roscoe, and I.J. Tree, Topology and its Applications $$39$$ ($$1991$$), $$71$$-$$103$$.
If $$X$$ is Hausdorff but not countably compact, there is an infinite closed discrete $$D\subseteq X$$; say $$D=\{x_n:n\in\Bbb N\}$$. For each $$n\in\Bbb N$$ there is an open $$U_n$$ such that $$U_n\cap D=\{x_n\}$$. For each $$m\in\Bbb N$$ let $$D_m=\{x_n\in D:2^m\le n<2^{m+1}\}$$; clearly $$|D_m|=2^m$$. Since $$D_m$$ is finite, the Hausdorffness of $$X$$ ensures that the points $$x_n\in Y_m$$ have pairwise disjoint open nbhds $$V_n$$, respectively, and we may further assume that $$V_n\subseteq U_n$$ for each $$n\in\Bbb N$$.
Now let $$\mathscr{V}_m=\{V_n:x_n\in D_m\}$$, and let $$\mathscr{V}=\{X\setminus D\}\cup\bigcup_{m\in\Bbb N}\mathscr{V}_m$$; clearly $$\mathscr{V}$$ is an open cover of $$X$$. Let $$F\subseteq X$$ be finite, and let $$m=|F|$$. Then $$m<2^m=|D_m|=|\mathscr{V}_m|$$, and each member of the pairwise disjoint family $$\mathscr{V}_m$$ can contain at most one point of $$F$$, so there is some $$n$$ such that $$x_n\in D_m$$ and $$V_n\cap F=\varnothing$$. But then $$x_n\notin\operatorname{st}(F,\mathscr{V})$$, so $$X$$ is not star compact.