Is every regular star compact metaLindelof space compact? A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.
Star compactness is stronger than pseudocompactness and weaker than countable compactness.
Is every regular star compact metaLindelof space compact?
Thanks for your help.
 A: Here I give a proof for the question. However I'm not sure that it is right. Could you help me ? Thanks for your any comment.
Proof: Let $X$ be a star compact metaLindelöf space. To prove that $X$
is compact it suffices to show that $X$ is Lindelöf since every
regular star compact Lindelöf space is compact.
Suppose not. Then there exists an open cover $\mathcal U$ of $X$
such that $\mathcal U$ contains no countable subcover. Since $X$ is
metaLindelöf, it follows that $\mathcal U$ has an
point-countable open refinement $\mathcal U'$. For $\mathcal U'$ as
an open cover of $X$ there exists a closed discrete set $D$ of $X$
such that $\operatorname{St}(D, \mathcal U') = X$. To see it, choose
inductively a point $x_\alpha \notin \operatorname{St}(\{x_\beta:
\beta < \alpha\}, \mathcal U')$. If $\lambda$ is the first ordinal
for which this choice is impossible, then $D=\{x_\alpha: \alpha <
\lambda \}$ is a closed discrete subset of $X$ and
$\operatorname{St}(D, \mathcal U') = X$. It is evident that $|D| >
\omega$ otherwise $\mathcal U$ would contain countable subcover.
Let $\mathcal W=\{\operatorname{St}(x_\alpha, \mathcal U'): x_\alpha
\in D\}$. Clearly, $\mathcal W$ is a new point-countable open cover
of $X$ the cardinality of which is greater than $\omega$. Since $X$
is star compact, it follows that there is a compact subset $K$ of
$X$ such that $\operatorname{St}(K, \mathcal W) = X$. It is not
difficult to see that, for each $x_\alpha \in D$, $K \cap
\operatorname{St}(x_\alpha, \mathcal U') \not= \emptyset$. Pick a
point $y_\alpha \in K \cap \operatorname{St}(x_\alpha, \mathcal U')$
and let $K'=\{y_\alpha: \alpha < \lambda\} \subset K$. It is clear
that $|K'| > \omega$ otherwise we can conclude that $|\mathcal W|
\le \omega$, which contradicts that $|\mathcal W| > \omega$.
Finally we prove that $K'$ is a closed discrete subset of $K$. To
see it, it suffices to show that for any $z \in X \setminus K'$,
there is a neighborhood $U$ of $z$ such that $U \cap K'=\emptyset$.
In fact, since $\mathcal W$ covers $X$, there exists a point
$x_\alpha \in D$ such that $z \in \operatorname{St}(x_\alpha,
\mathcal U')$. Choose an open set $U$ such that $y_\alpha \notin U$
and $U \subset \operatorname{St}(x_\alpha, \mathcal U')$. Clearly,
$U \cap K' = \emptyset$.
Now we can conclude that $K'$ is an infinite closed discrete subset
of $K$. This contradicts that $K$ is compact. This completes the
proof.
A: $\newcommand{\st}{\operatorname{st}}$You can also derive the result as a corollary of the result that a $T_1$, star-Lindelöf, metaLindelöf space is Lindelöf, which can be proved using the same ideas, but perhaps even more easily.
Suppose that $X$ is $T_1$, star-Lindelöf and metaLindelöf. Let $\mathscr{U}$ be an open cover of $X$; $\mathscr{U}$ has a point-countable open refinement $\mathscr{V}$, and there is a Lindelöf $A\subseteq X$ such that $\st(A,\mathscr{V})=X$. Recursively construct a set $A_0=\{x_\xi:\xi<\alpha\}\subseteq A$ such that $\bigcup_{\xi<\alpha}\st(x_\xi,\mathscr{V})=X$ and $x_\eta\notin\st(x_\xi,\mathscr{V})$ whenever $\xi<\eta<\alpha$. For $\xi<\alpha$ let $W_\xi=\st(x_\xi,\mathscr{V})$; $W_\xi\cap A_0=\{x_\xi\}$, so $A_0$ is a closed, discrete subset of $X$. But then $A_0$ is a closed subset of the Lindelöf subspace $A$, so $A_0$ is Lindelöf and must therefore be countable, $\{V\in\mathscr{V}:V\cap A_0\ne\varnothing\}$ is a countable subcover of $\mathscr{V}$, and $X$ is Lindelöf.
