TITLE: What fragment of ZFC do we need to prove Zorn's lemma?
QUESTION [14 upvotes]: It is extremely well-known that Zorn's lemma is a theorem of ZFC. My interest is in a certain finitely-axiomatisable fragment of ZFC, sometimes called RZC (restricted Zermelo with choice) or ZBQC. The axioms of RZC are the following:

Extensionality
Empty set
Pair set
Union
Power set
Infinity
Foundation
Choice (in the sense that every surjection has a right inverse)
Separation for $\Delta_0$-formulae

Adrian Mathias also defines an extension of RZC, called MAC (Mac Lane set theory), by adding the transitive containment axiom:

Every set is contained in a transitive set.

(Apparently RZC and MAC are equiconsistent.)
Question. Is Zorn's lemma a theorem of RZC? If not, is it a theorem of MAC?
It is reasonably clear that well-founded induction is valid in RZC, but without the axiom of replacement it is not at all obvious to me whether Hartogs numbers exist. ($V_{\omega + \omega}$ is a model of RZC, but the only von Neumann ordinals in $V_{\omega + \omega}$ are precisely those below $\omega + \omega$, even though it has uncountable well-ordered sets.) Once we know that there are sufficiently large well-ordered sets, it seems to me that the usual proof of Zorn's lemma will go through in RZC.
Motivation. One can build a model of RZC out of any model of ETCS (elementary theory of the category of sets) and ETCS can be interpreted in any model of RZC. What I really want to know is whether ETCS proves that, say, every vector space has a basis, and it seems like a good first step would be to establish the claim for RZC.

REPLY [9 votes]: Actually, one does not need to use any ordinals to prove Zorn's lemma in RZC!
Theorem: Every non-empty partial order $(S,<)$ in which every non-empty chain has an upper bound has a maximal element.
Proof: We shall work under the assumption that the claim is false. First let $F$ be a choice-function that maps each chain $C$ in $(S,<)$ to a strict upper bound for $C$ in $S$. (This is the only use of AC in this proof, and also shows that it suffices for $S$ to be well-orderable in absence of AC.) We say that $T$ is a tower iff $T$ is a well-ordered chain in $(S,<)$ and $∀x{∈}T\ ( \ x = F(T_{<x}) \ )$. Note that any two towers $T,U$ agree, meaning that one is a subchain of the other, since otherwise $T{∖}U$ and $U{∖}T$ are both non-empty and disjoint and so letting $m = \min_<(T{∖}U)$ and $n = \min_<(U{∖}T)$ we would obtain $m = F(T_{<m}) = F(U_{<n}) = n$ and hence a contradiction. Now let $V$ be the union of all the towers, and observe by the above note that it is a tower too, but that $V∪{F(V)}$ is a taller tower, contradicting the definition of $V$.