I was looking into the proof of the Ultrafilter lemma by Zorn's lemma, and I wasn't able to find a sufficient explanation for the following claim:
If $\mathcal{C}\subseteq P$ is a chain of proper filters, then $F:=\cup \mathcal{C}$ is also a proper filter.
I'm okay with showing that $F$ is a filter, but it's not clear to me why it is proper.
Recall that a filter $\cal F$ is proper if and only if $\varnothing\notin\cal F$.
Now suppose that $\cal F$ is the union of any family of filters, $\bigcup\cal C$. If $\varnothing\in\cal F$, then by the definition of $\bigcup\cal C$, there is some $X\in\cal C$ such that $\varnothing\in X$. Therefore $X$ could not have been a proper filter.