Proposition 1   Kuratowski's lemma implies Zorn's lemma.

Proof. Suppose $P$ is a poset such that every chain has an upper bound. Let $C$ be a chain in $P$ . By Kuratowski's lemma, $C$ can be extended to a maximal chain $C'$ , which, by assumption, has an upper bound $a\in P$ . Suppose for some $b\in P$ , $a\le b$ . If $b\ne a$ , then $b\not\le a$ , or $b\notin C'$ , which means $C'\cup \lbrace b\rbrace$ is a chain extending $C'$ , thus extending $C$ . But this means that $C'$ is not maximal, a contradiction. This shows that $b=a$ , or that $a$ is a maximal element in $P$ , proving Zorn's lemma.

Proposition 2   Zorn's lemma implies Kuratowski's lemma.

Proof. Suppose $P$ is a poset and $C$ a chain in $P$ . We assume that $P\ne \varnothing$ . Let $\mathcal{P}$ be the set of all chains in $P$ extending $C$ . Partially order $\mathcal{P}$ by inclusion so that $\mathcal{P}$ is a poset. Let $\mathcal{C}$ be a chain in $\mathcal{P}$ . Let $C'=\bigcup \mathcal{C}$ . We want to prove that $C'$ is an upper bound of $\mathcal{C}$ in $\mathcal{P}$ .

1. $C'$ is a chain in $P$ . If $x,y\in C'$ , then $x\in D$ and $y\in E$ for some $D,E\in \mathcal{C}$ . Since $\mathcal{C}$ is a chain in $\mathcal{P}$ , $D\subseteq E$ or $E\subseteq D$ , which implies that $x,y$ belong to the same chain (either $D$ or $E$ ) in $P$ . So $x\le y$ or $y\le x$ . This shows that $C'$ is a chain in $P$ .
2. $C'$ is in $\mathcal{P}$ . If $a\in C$ , then $a\in D$ for every $D\in \mathcal{C}$ , and therefore $a\in C'$ , showing that $C'$ extends $C$ , or $C' \in \mathcal{P}$ .
3. $C'$ is an upper bound of $\mathcal{C}$ . Pick any $x\in D$ , for an arbitrary $D\in \mathcal{C}$ . Then $x\in \bigcup \mathcal{C}=C'$ , so $D\subseteq C'$ . Since $D$ is arbitrary, $C'$ is an upper bound of $\mathcal{C}$ .

By Zorn's lemma, $\mathcal{P}$ has a maximal element $M$ . Then $M$ is a maximal chain in $P$ extending $C$ , for if there is a chain $N$ in $P$ such that $M\subseteq N$ , then $N\in \mathcal{P}$ and $M$ would no longer be maximal.