How far is star-compactness from countably compactness?
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 $X = \operatorname{St}(K,\mathscr{U})$.
I've found that every regular $T_1$ star compactness space $X$ is countably compact, which leads a paradox. However, I don't know where I'm wrong.
Prove: In every regular $T_1$ space $X$, countably compactness is equiavalent to this condition: every point finite countable open covering of $X$ has a finite subcover (see the theorem 2 of Pseudocompact and Countably Compact Spaces). For any point finite countable open covering $\mathscr{U}$ of $X$, there is a compact subspace $K$ of $X$ such that $St(K,\mathscr{U})=X.$ Suppose $X$ is not coutably compact, then $\mathscr{U}$ has not finit subcover. For any $x_1 \in K$, there exists a finit subcollection $\mathscr{U_1}$ of $\mathscr{U}$, such that $x_1$ is in the every element of $\mathscr{U_1}$ and $\cup \mathscr{U_1}$ can't cover $X$. So there exists $x_2 \in K\setminus \cup \mathscr{U_1}$ and $\mathscr{U_2}$, such that $x_2$ in the every element of $\mathscr{U_2}$ and $\cup \{\mathscr{U_i}, i= 1,2\}$ can't cover $X$. We can work on it until $\omega$ times! Therefore there exists a subset $\{x_i:i \in \omega\}$ of $K$ which is a countable closed discrete set of the compact $K$. Contradiction!