proof of Tukey’s lemma

Let $S$ be a set and $F$ a set of subsets of $S$ such that $F$ isof 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 orderedby 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 andthe $F_i$ 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.