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The proofs I know of Zorn's lemma give the following refinement:

Let $(X,<)$ be a partially ordered set such that every well-ordered subset of $X$ has an upperbound. Then $X$ has a maximal element.

In fact, Zorn's lemma is sometimes stated as such. In comparison, the usual statement asks that every totally ordered subset of $X$ has an upper bound.

Does this refinement have some application?

EDIT (after comments). Your comments make me realize that I did not think enough of my question, which is what the voters-to-close probably guessed. To help future readers, let me sum up the comments.

  • If $(X,<)$ satisfies the hypothesis of the refined Zorn lemma (RZL), then (modulo AC), it satisfies the hypothesis of classical Zorn lemma as well, since any totally ordered set contains a cofinal ordered subset. (Comment of Ramiro de la Vega).

  • ZF+RZL implies ZF+AC (Noah S).

  • In Cohen's first model of ZF, there exists a set which satisfies the hypothesis of RZL but not that of ZL (Asaf Karagila).

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    $\begingroup$ Since that seemingly weaker version of Zorn implies AC, they are in fact equivalent. (To see that this version implies AC: given a set $A$ consider the poset of partial well-orderings of $A$.) $\endgroup$
    – Noah Schweber
    Oct 16, 2014 at 16:22
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    $\begingroup$ @NoahS: I believe the question asks for examples where checking only well-ordered chains is much easier than checking all chains. $\endgroup$
    – Ramiro de la Vega
    Oct 16, 2014 at 16:25
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    $\begingroup$ One interesting way to interpret the question is to ask for a model of ZF in which AC fails, which has a partial order in which all well-ordered chains have upper bounds, but some totally ordered chains do not. This would show, if indeed there are such models, that the equivalence of the two hypotheses was itself a kind of choice principle. $\endgroup$
    – Joel David Hamkins
    Oct 16, 2014 at 17:37
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    $\begingroup$ @Noah: Cohen's first model satisfies that. The Dedekind-finite set of reals is unbounded, but every well-ordered chain is finite, so it has a maximal element. $\endgroup$
    – Asaf Karagila ♦
    Oct 16, 2014 at 18:15
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    $\begingroup$ Can one of the voters-to-close explain why they're voting to close? $\endgroup$
    – Tom Leinster
    Oct 16, 2014 at 18:26

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According to Campbell, Paul J. The origin of "Zorn's lemma''. (French summary) Historia Math. 5 (1978), no. 1, 77–89, results in topology gave motivation to the Kuratowski-Zorn lemma (as it is known where I come from), starting with the result of Zygmunt Janiszewski from 1910 that every continuum between two given points contains a minimal continuum. The version of the lemma highlighted in your question was in fact proved by Kazimierz Kuratowski in 1922 (that is, 13 years before Max Zorn came up with his), in the paper. "Une méthode d'élimination des nombres transfinis des raisonnements mathématiques", Fundamenta Mathematicae 3: 76–108. http://matwbn.icm.edu.pl/ksiazki/fm/fm3/fm3114.pdf As applications of his theorem, Kuratowski gives the proofs of the mentioned result by Janiszewski (in a generalized form), as well as other theorems in point-set topology (e.g. Cantor-Bendixson theorem), Lebesgue's theorem on the existence of an uncountable class of Borel measurable sets and the existence of some Baire classes of functions.

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