If a topological space has $\aleph_1$-calibre and cardinality at most $2^{\aleph_0}$ must it be star-countable? If a topological space $X$ has $\aleph_1$-calibre and the cardinality of $X$ is $\le 2^{\aleph_0}$, then it must be star countable? A topological space $X$ is said to be star-countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $A$ of $X$ such that $X = \operatorname{St}(A,\mathscr{U})$.

Under CH the space is separable (hence, star countable ). 
Proof(Ofelia). On the contrary, suppose that X is not separable .Under CH write $X = \{ x_\alpha : \alpha \in \omega_1 \}$ and for each $\alpha$ in $\omega_1$ define $U_\alpha$ = the complement of $cl ( { x_\beta : \beta \le \alpha } )$ . The family of the $U_\alpha$ is a decreasing family of non-empty open sets; since $\aleph_1$ is a caliber of X, the intersection of all $U_\alpha$ must be non-empty (contradiction!)
