Suppose that $X$ is a set and $\mathcal A$ is a collection of subsets of $X$ such that $X = \bigcup \mathcal{A}$ and for every countable $\mathcal{A}_0\subset \mathcal{A}$ we have $X\neq \bigcup \mathcal{A}_0$. Is it possible to find a subfamily $\mathcal{B}\subset \mathcal{A}$ such that the family $\mathcal{C}:=(\mathcal{A}\setminus \mathcal{B}) \cup \{ \bigcup \mathcal{B} \}$
- has cardinality $\aleph_1$,
- $X = \bigcup \mathcal C$, and
- for every countable $\mathcal C_0\subset \mathcal C$ we have $X\neq \bigcup \mathcal C_0$?
No, not necessarily. For instance, let $X=\omega_2$ and let $\mathcal{A}=\omega_2$ as well (that is, $\mathcal{A}$ is the set of proper initial segments of $X$, which are all the ordinals less than $\omega_2$). If $\mathcal{B}\subseteq\mathcal{A}$ is unbounded, then $\bigcup \mathcal{B}=X$ and so $\mathcal{C}$ will not work (there is a single element of it that already covers $X$). If $\mathcal{B}\subseteq\mathcal{A}$ is bounded, then it is bounded by an ordinal $\alpha$ of cardinality $\aleph_1$, so $\mathcal{C}=(\mathcal{A}\setminus\mathcal{B})\cup\{\bigcup\mathcal{B}\}\supseteq\omega_2\setminus\alpha$ has cardinality $\aleph_2$.