Let $S$ be a set and $F$ a set of subsets of $S$ such that $F$ is of finite character. By Zorn's lemma, it is enough to show that $F$ is inductive. For that, it will be enough to show that if $(F_i)_{i\in I}$ is a family of elements of $F$ which is totally ordered by inclusion, then the union $U$ of the $F_i$ is an element of $F$ as well (since $U$ is an upper bound on the family $(F_i)$ . So, let $K$ be a finite subset of $U$ Each element of $U$ is in $F_i$ for some $i\in I$ Since $K$ is finite and the $F_i$ are totally ordered by inclusion, there is some $j\in I$ such that all elements of $K$ are in $F_j$ That is, $K\subset F_j$ Since $F$ is of finite character, we get $K\in F$ QED.