I'm currently reading Dugundji's Topology book. Exercise 2 of section 2 in page 58 gives 4 equivalents of Zorn's lemma. I'm only interested in a) and c) (I won't copy the whole thing cause it's too long but many people will surely have the book):
a) [H. Kneser's form of Zorn's Lemma.] If X is a preordered set, and if each well-ordered subset has an upper bound, then X has at least one maximal element.
c) If X is a partially ordered set, then each chain in X is conatined in a maximal chain. [M is maximal chain if ($M\subseteq C)\implies (M=C)$ for each chain $C\subseteq X$.]
Dugundji's book suggest to prove a) by modifying the argument in the text. I don't know how to modify the argument, one surely has to modify the definition of median but I can't get it. I proved only AC then c) and a) then Zorn (or AC) is clear so I need the other two implications.
I’ve left out a few details.
Assume (c), and let $\mathscr{A}=\{A_i:i\in I\}$ be a family of non-empty sets. Let $F$ be the set of functions $f:J\to\bigcup\mathscr{A}$ such that $J\subseteq I$, and $f(j)\in A_j$ for each $j\in J$. (In other words, $F$ is the set of partial choice functions for $\mathscr{A}$. Each $f\in F$ is a subset of $I\times\bigcup\mathscr{A}$, so $\langle F,\subseteq\rangle$ is a partial order. Fix $i_0\in I$ and $a_0\in A_{i_0}$, and let $f_0=\{\langle i_0,a_{i_0}\rangle\}\in F$; clearly $\{f_0\}$ is a chain in $\langle F,\subseteq\rangle$, so it’s contained in a maximal chain $M$. Let $m=\bigcup M$; then $m\in F$. If $\operatorname{dom}m\ne I$, fix $i_1\in I\setminus\operatorname{dom}m$ and $a_1\in A_{i_1}$; then $m^+=m\cup\{\langle i_1,a_{i_1}\rangle\in F$, and $M\cup\{m^+\}$ is a chain properly extending $M$, contradicting the maximality of $M$. Thus, $m$ is a choice function for $\mathscr{A}$, and (c) implies $\mathsf{AC}$.
Now assume Zorn’s lemma, and let $\langle X,\le\rangle$ be a preorder in which every well-ordered subset has an upper bound. Let $\mathscr{W}$ be the family of all well-ordered subsets of $X$. Define a relation $\preceq$ on $\mathscr{W}$ by $W_0\preceq W_1$ if and only if $\langle W_0,\le\rangle$ is an initial segment of $\langle W_1,\le\rangle$. It’s not hard to check that $\preceq$ is a partial order. Suppose that $\mathscr{C}$ is a chain in $\mathscr{W}$. Show that $\bigcup\mathscr{C}\in\mathscr{W}$, and that $C\preceq\bigcup\mathscr{C}$ for each $C\in\mathscr{C}$. Thus, $\bigcup\mathscr{C}$ is an upper bound for $\mathscr{C}$, so by Zorn’s lemma $\mathscr{W}$ has a maximal element $W$. By hypothesis $W$ has an upper bound $w$. If $w<x$ for some $x\in X$, then $W\prec W\cup\{x\}\in\mathscr{W}$, contradicting the maximality of $W$, so $w$ must be maximal in $\langle X,\le\rangle$. (And incidentally we must have $w\in W$.)