Zorn's lemma

Suppose $P$ has no maximal element.

Let $\mathcal{C}$ be the set of all chains of $P$, a subset of the power set of $P$, partially ordered by inclusion.

$(\mathcal{C}, \subseteq)$ is a chain-complete poset: if $\mathcal{D} \subseteq \mathcal{C}$ is a chain under inclusion then $\bigcup \mathcal{D}$ is a chain of $P$, since any two of its elements lie in a common member of $\mathcal{D}$, and it is the least upper bound of $\mathcal{D}$ under inclusion; the empty chain has least upper bound $\bigcup \emptyset = \emptyset$, which is a chain.

For $C \in \mathcal{C}$ let $S_C$ be the set of strict upper bounds of $C$ in $P$.

Each $S_C$ is nonempty: $C$ has an upper bound $u$ in $P$ by hypothesis, taking any element of the nonempty $P$ when $C = \emptyset$; by assumption $u$ is not maximal, so there is $v$ with $u < v$; then for every $c \in C$ transitivity gives $c \le v$ from $c \le u \le v$, and $c \ne v$, since $c = v$ would give $v \le u$ and $u \le v$, hence $u = v$ by antisymmetry, contradicting $u < v$; so $c < v$ for every $c \in C$ and $v \in S_C$.

Apply the Axiom of Choice to the family $\mathcal{S} = \{S_C : C \in \mathcal{C}\}$, every member of which is nonempty, obtaining a choice function $g$ with $g(S) \in S$ for each $S \in \mathcal{S}$; composing with the map $C \mapsto S_C$, which is a function on $\mathcal{C}$, yields a selection $C \mapsto g(S_C) \in S_C$ defined for every chain $C$, and no injectivity of $C \mapsto S_C$ is needed, since two chains with the same set of strict upper bounds simply receive the same chosen element.

Define $f(C) = C \cup \{g(S_C)\}$ for $C \in \mathcal{C}$; this is again a chain, because $g(S_C)$ is a strict upper bound of $C$ and so is comparable to every element of $C$.

$f$ is progressive for inclusion, since $C \subseteq f(C)$ by construction.

By Bourbaki–Witt applied to the chain-complete $\mathcal{C}$ and the progressive $f$, there is $C_0 \in \mathcal{C}$ with $f(C_0) = C_0$, that is $g(S_{C_0}) \in C_0$.

But $g(S_{C_0})$ is a strict upper bound of $C_0$, so every element of $C_0$ is strictly below it, giving $g(S_{C_0}) < g(S_{C_0})$, which is impossible because $<$ is irreflexive.