What properties of $\omega$-covers also hold for any open cover?

Question

As a continuation to this interesting question.

Suppose that every $\omega$-cover of $X$ has a countable subcover which is also an $\omega$-cover. Does it implies that every cover of $X$ has a countable subcover?

Thank you!

Answer 1

Let $X$ be a space such that every $\omega$-cover has a countable subfamily which is also an $\omega$-cover. Fix an open cover $\mathcal{U}$ of $X$. Then the family $\mathcal{V}$ of all finite unions of sets in $\mathcal{U}$ is an $\omega$-cover of $X$, so there are $\{ V_i : i \in \mathbb{N} \} \subseteq \mathcal{V}$ which is also an $\omega$-cover of $X$. For $i \in \mathbb{N}$, let $U_{i,1} , \ldots , U_{i,n_i} \in \mathcal{U}$ be such that $V_i = U_{i,1} \cup \cdots \cup U_{i,n_i}$. It easily follows that $\{ U_{i,j} : i \in \mathbb{N} , j \leq n_i \}$ is a countable subcover of $\mathcal{U}$.