As a continuation to this question:
Given the space of countable complement topology on $X$, where $X$ is an uncountable set. (example 20 in "Counterexamples in topology"). We know that $X$ is not $\sigma$. My question is: Can I assume that $X$ has property $S_1(\mathcal O,\mathcal O)$
Definition: Let $\langle \mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of open covers of $X$. If we can always find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each $F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an open cover of $X$, we say that $X$ has property $S_1(\mathcal O,\mathcal O)$.
Thank you!
Choose one open set in $\mathcal{U}_1$. There are countably many points not in that open set. Enuerate them and cover them one at a time with an open set from $\mathcal{U}_n$. This gives you an open cover.