(Already presented in the Well-Ordering Theorem page.) Given AC, fix a choice function $f$ on $\mathcal{P}(A) \setminus \{\emptyset\}$, and use transfinite recursion to enumerate $A$. $\blacksquare$

Let $(P, \leq)$ be a poset where every chain has an upper bound. By WO, well-order $P$ as $\{p_\alpha : \alpha < \theta\}$.

Build a chain $C$ in $P$ by transfinite recursion:

• Start with $C_0 = \{p_0\}$.
• At stage $\alpha$: if $p_\alpha$ is an upper bound for $C_{<\alpha} = \bigcup_{\beta < \alpha} C_\beta$ and $p_\alpha \geq c$ for all $c \in C_{<\alpha}$, add $p_\alpha$ to the chain: $C_\alpha = C_{<\alpha} \cup \{p_\alpha\}$.
• Otherwise, skip $p_\alpha$: $C_\alpha = C_{<\alpha}$.

More carefully: define $C$ as a maximal chain. Well-order $P$ by $\prec$. Define by recursion: $c_0$ is the $\prec$-least element of $P$. Given the chain $\{c_\beta : \beta < \alpha\}$, let $U_\alpha$ be the set of upper bounds in $P$. If $U_\alpha \neq \emptyset$, let $c_\alpha$ be the $\prec$-least element of $U_\alpha$ with $c_\alpha \geq c_\beta$ for all $\beta < \alpha$ (or more precisely, choose $c_\alpha$ to extend the chain).

If the chain $C$ has no strict upper bound (i.e., no element of $P$ is strictly greater than all elements of $C$), then an upper bound $m$ of $C$ (which exists by hypothesis) satisfies: $m$ is maximal. For if $m \leq p$, then $p$ would be an upper bound of $C$ with $p \geq m$. If $p > m$, we could have added $p$ to the chain, contradicting the construction. Hence $p = m$, and $m$ is maximal. $\blacksquare$

Let $\{A_i\}_{i \in I}$ be a family of nonempty sets. Define the poset:

$P = \left\{f : f \text{ is a function}, \mathrm{dom}(f) \subseteq I, \text{ and } f(i) \in A_i \text{ for all } i \in \mathrm{dom}(f)\right\},$

ordered by inclusion (extension): $f \leq g$ iff $\mathrm{dom}(f) \subseteq \mathrm{dom}(g)$ and $g|_{\mathrm{dom}(f)} = f$.

$P$ is nonempty: $\emptyset \in P$.

Every chain $\{f_\alpha\}_{\alpha \in J}$ in $P$ has an upper bound: $g = \bigcup_\alpha f_\alpha$ is a well-defined function (compatibility follows from the chain condition), $\mathrm{dom}(g) = \bigcup \mathrm{dom}(f_\alpha) \subseteq I$, and $g(i) \in A_i$ for each $i \in \mathrm{dom}(g)$.

By Zorn's lemma, $P$ has a maximal element $f^*$. If $\mathrm{dom}(f^*) \neq I$, pick $i_0 \in I \setminus \mathrm{dom}(f^*)$ and any $a \in A_{i_0}$ (nonempty), then $f^* \cup \{(i_0, a)\}$ strictly extends $f^*$, contradicting maximality. So $\mathrm{dom}(f^*) = I$, and $f^*$ is the desired choice function. $\blacksquare$