Tukey finite character is equivalent to AC

Assume AC and let $\mathcal{F}$ have finite character. If $Y\in \mathcal{F}$, every subset of $Y$ belongs to $\mathcal{F}$, since its finite subsets are finite subsets of $Y$. In particular $\mathrm{\varnothing }\in \mathcal{F}$. For a nonempty inclusion-chain $\mathcal{C}\subseteq \mathcal{F}$, every finite subset of $\bigcup \mathcal{C}$ is contained in one chain member: choose finitely many covering members and take the largest among them. Thus $\bigcup \mathcal{C}\in \mathcal{F}$. The empty chain has upper bound $\mathrm{\varnothing }$.

Assume Tukey and let $(X_i{)}_{i\in I}$ be any nonempty-set family. Inside $I\times {\bigcup }_i{X}_i$ let $\mathcal{G}$ consist of graphs of partial functions $g$ satisfying $g(i)\in X_i$. It contains the empty graph. A graph fails the conditions only by a bad pair $(i,x)$ with $x\mathrm{\notin }{X}_i$, or by two pairs with the same first coordinate and different values. These witnesses have sizes one and two, so $\mathcal{G}$ has finite character.

Apply Zorn to $(\mathcal{F},\subseteq )$ to obtain an inclusion-maximal member.

A maximal $g\in \mathcal{G}$ must have domain $I$: at an omitted $i$, any one $x\in X_i$ extends it, contradicting maximality. If $I$ is empty the empty graph already suffices. Thus every family has a choice function.