Let $X=\omega_2$ and $\tau=\{U\subset X: |X\setminus U|<\omega_2\}$ is the topology on $X$. Then could the space $X$ be star countable?
If it is not, what about the space $Z=\omega_1$ and $\tau'=\{U\subset X: |X\setminus U|<\omega_1\}$?
Thanks for any help.
Both spaces are star countable.
More generally, let $\kappa$ be an infinite regular cardinal, let $\mathscr{F}=\{F\subseteq\kappa:|F|<\kappa\}$, and let $$\tau=\{U\subseteq\kappa:\kappa\setminus U\in\mathscr{F}\}\cup\{\varnothing\}\;;$$ then $\tau$ is a star countable topology on $\kappa$.
To see this, let $\mathscr{U}$ be an open cover of $\kappa$. Fix $V\in\mathscr{U}$. For each $\xi\in\kappa\setminus V$ there is a $V_\xi\in\mathscr{U}$ such that $\xi\in V_\xi$. Let $\mathscr{V}=\{V\}\cup\{V_\xi:\xi\in\kappa\setminus V\}\subseteq\mathscr{U}$; clearly $\mathscr{V}$ covers $\kappa$. Let $F=\bigcup_{W\in\mathscr{V}}(\kappa\setminus W)$; then $F$ is the union of fewer than $\kappa$ sets, each of cardinality less than $\kappa$, so $|F|<\kappa$, and
$$\bigcap\mathscr{V}=\kappa\setminus F\ne\varnothing\;.$$
Thus, $X=\operatorname{St}\big(\{\xi\},\mathscr{V}\big)$ for each $\xi\in\bigcap\mathscr{W}$, and $\kappa$ is not just star countable, but star-$1$, so to speak.