For each subset of $A\subseteq S$ which is well-ordered under the preorder it inherits from $S$, choose a strict upper bound $f(A)\in S$ if it has one. (Note that we use the axiom of choice here. We also freely use its consequence, the axiom of excluded middle, below.)

Say a well-ordered subset $A$ is $f$-inductive if for all $x\in A$, $x=f(\{y\in A:y<x\})$. Thus $A_0=\varnothing$ is inductive, as is $A_1=\{f(A_0)\}$, $A_2=A_1\cup \{f(A_1)\}$, $A_3=A_2\cup \{f(A_2)\}$ and so on. Intuitively, using the chain hypothesis, one can (by transfinite induction) define $A_{\alpha }$ for ordinals $\alpha$ by appending strict upper bounds in this way, until at some point $\alpha$, one has run out of further strict upper bounds to choose from. But by hypothesis, $A_{\alpha }$ does have an upper bound $x$. This element is necessarily maximal.

More precisely, $f$-inductive sets are well-ordered by inclusion. For if $A$ and $B$ are distinct (well-ordered and) $f$-inductive sets, then there is a smallest element contained in one but not the other, say $x\in A$ but $\neg (x\in B)$. But then $\{y\in A:y<x\}$ is a down-closed subset of $B$, i.e., is an initial segment of $B$. If it were a proper subset of $B$, then for the least $z\in B$ in its complement we would have

whence by applying $f$ to both sides, we infer $x=z$ by $f$-inductivity of $A$ and $B$, contradiction. In that case, $B=\{y\in A:y<x\}$. In other words, one of $A,B$ must be an initial segment of the other.

Therefore the union of all $f$-inductive sets is also $f$-inductive. Being the maximal $f$-inductive set $M$, it contains any of its upper bounds (in fact there is only one up to isomorphism), and this element is necessarily maximal in $S$.

Let $A$ be a set, and consider the poset whose elements are pairs $(X,R)$, where $X\subseteq A$ and $R$ is a (classical) well-ordering on $X$, ordered by $(X,R)\le (Y,S)$ if $X$ as a well-ordered set is an initial segment of $Y$. The hypothesis of Zorn’s lemma is easily established: given a chain $(X_t,R_t{)}_{t\in T}$, put $X={\bigcup }_t{X}_t$ and define the relation $R$ on $X$ by $xRy$ if $x{R}_{t}y$ for any (hence all) $X_t$ containing both $x,y$. Then $R$ is a well-ordering: if $U\subseteq X$ is any nonempty set, then $U\cap X_t$ is nonempty for some $t$, and the minimal element of $U$ with respect to $R$ will also be the minimal element of $U\cap X_t$ with respect to $R_t$ (crucially using here the initial segment condition).

By Zorn’s lemma, the poset above contains a maximal element $(X,R)$, where $X\subseteq A$. We claim $X=A$; suppose not (here is where we need excluded middle), and let $x$ be a member of the complement $\neg X$. Well-order $X\cup \{x\}$, extending $R$ by deeming $x$ to be the final element. This shows $(X,R)$ was not maximal, contradiction. Hence any set $A$ may be well-ordered.

The axiom of choice follows: suppose given a surjection $p:E\to B$, so that every fiber $p^{-1}(b)$ is inhabited. Consider any well-ordering of $E$, and define $s:B\to E$ by letting $s(b)$ be the least element in $p^{-1}(b)$ with respect to the well-ordering. This gives a section $s$ of $p$.