Let $\mathcal{F}\neq \varnothing$ be of finite character and consider the strict partial order $\subsetneq$ in it. Let $C\subseteq \mathcal{F}$ be a chain and $A:=\bigcup C$.
If we prove that every finite subset of $A$ belongs to $\mathcal{F}$, then $A\in \mathcal{F}$. As every element of $C$ is a subset of $A$, then $A$ is an upper bound of $C$. Hence, by Zorn's Lemma, there exists a $\subsetneq$-maximal $B\in C$. QED
My doubt is: how do we prove that every finite subset of $A$ belongs to $\mathcal{F}$?
I've conjectured that "If $C$ is a non-empty set such that $\forall X,Y\in C (X\subseteq Y \vee Y\subseteq X)$, then $\bigcup C\in C$", but I couldn't prove that this is true.
I tried, then, to split the proof in the cases where $A\in C$ and the ones where $A\not\in C$.
If $A\in C$, then $A\in \mathcal{F}$, because $C\subseteq \mathcal{F}$.
If $A\not \in C$, then $C$ doesn't have a maximal element. But, by definitions, it just means that $\forall X\in C \exists Y\in C (X\subsetneq Y)$, so I didn't get anywhere.
I've read Asaf Karagila's answer to a quite similar question in the link below:
Confusion in Proof of "Zorn's Lemma $\implies$ Tukey's Lemma"
The answer is this one:
"Either $C$ has a maximal element, in which it's $A$. Or it doesn't, in which case every finite subset is a subset of some element of the chain."
The part that I don't get is this one: "Or it doesn't, in which case every finite subset is a subset of some element of the chain". Why is this true?
In case you wanna know, I'm using Kenneth Kunnen's "The Foundations of Mathematics" book. For the proof of this theorem he only gives a hint, but it didn't help me a bit.
Best Answer
Let $\{a_1,\ldots, a_n\}$ be a finite subset of $A:=\bigcup C$ and let $Z_1,\ldots, Z_n\in C$ such that $a_i\in Z_i.$ Since $C$ is a chain, the $Z_i$ are totally ordered by $\subseteq,$ so, reindexing if necessary, we can assume $Z_1\subseteq Z_2\subseteq\ldots \subseteq Z_n.$ Thus, for all $i\le n,$ $a_i\in Z_n,$ i.e. $\{a_1,\ldots, a_n\}\subseteq Z_n.$ Since $\mathcal F$ is of finite character and $Z_n\in \mathcal F$, $\{a_1,\ldots,a_n\}\in\mathcal F.$