A question on star countable space Here is a proposition from the paper of L.P. Aiken Star-covering properties. I cannot understand why (the last line) the space is star countable. Could somebody help me? Thanks ahead.
The picture is not clear. You could see the Proposition 11 of the paper.

Proposition 11. If $\aleph_0\leq\kappa\leq\mathfrak{c}$, there exists a m.a.d.f. $\mathcal{E}\subseteq [\kappa]^\omega$ such that $\Psi(\mathcal{E})$ is star-countable.
Proof. Let $\mathcal{C}\subseteq[\omega]^\omega$ and $\mathcal{D}\subseteq[\kappa\setminus \omega]^\omega$ be maximal almost disjoint families such that $|\mathcal{C}|=|\mathcal{D}|=\mathfrak{c}$. This is possible because $\kappa\leq\mathfrak{c}$. Choose a bijection $f:\mathcal{C}\to\mathcal{D}$. Define $\mathcal{E}=\{C\cup f(C):C\in\mathcal{C}\}$. If $A\in[\kappa]^\omega$, $|A\cap\omega|=\aleph_0$ or $|A\cap (\kappa\setminus\omega)|=\aleph_0$. In either case, $A$ has infinite intersection with some element of $\mathcal{C}\cup\mathcal{D}$, thus $\mathcal{E}$ is maximal. Then $\Psi(\mathcal{E})$ is star-countable because $\mathsf{St}(\omega,\mathscr{U})=\Psi(\mathcal{E})$, for any open cover $\mathscr{U}$ of $\Psi(\mathcal{E})$. $\;\;\Box$

(original image)
 A: If $\mathscr{U}$ is any open cover of $\Psi ( \mathcal{E} )$, it suffices to show $\Psi ( \mathcal{E} ) \setminus \mathrm{st} ( \omega , \mathscr{U} )$ is finite.


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*Fist note that every $E \in \mathcal{E}$ belongs to $\mathrm{st} ( \omega , \mathscr{U} )$.  Pick any $U \in \mathscr{U}$ containing $E$.  Then in particular $E \setminus U$ is finite.  Note that $E$ itself is of the form $C \cup f (C)$ where $C \in \mathcal{C}$, and as $C \subseteq \omega$ is infinite it must be that $U \cap \omega$ is infinite, in particular it is nonempty.  Therefore $E \in \mathrm{st} ( \omega , \mathscr{U} )$.

*It follows that $Z = \Psi ( \mathcal{E} ) \setminus \mathrm{st} ( \omega , \mathscr{U} ) \subseteq \kappa$.  I claim that $Z \cap E$ is finite for all $E \in \mathcal{E}$.  Given $E \in \mathcal{E}$ there is a $U \in \mathscr{U}$ such that $E \in U$.  It follows that $E \setminus U$ is finite, and $U \subseteq \mathrm{st} ( \omega , \mathscr{U} )$ (by the observation in the first point).  Therefore $Z \cap U = \emptyset$ and so $Z \cap E \subseteq E \setminus U$ is finite.  Since $\mathcal{E}$ is a m.a.d.f. it follows that $Z$ cannot be infinite.

Note that the proof given in the paper is incorrect since given $\alpha \in \kappa \setminus \omega$ the family $$\mathscr{U} = \{ \{ E \} \cup ( E \setminus \{ \alpha \} ) : E \in \mathcal{E} \} \cup \{ \alpha \}$$ is an open cover of $\Psi ( \mathcal{E} )$ but $\alpha \notin \mathrm{st} ( \omega , \mathscr{U} )$.  (Since $\mathcal{E}$ is a m.a.d.f. we may assume without loss of generality that $\bigcup \mathcal{E} = \kappa$.)
