I was looking through Set Theory and the Continuum Problem by Smullyan and Fitting and came across this note in Exercise 1.2 in Chapter 11 (pg 147 of Dover version).
(We remark that the Skolem-Löwenheim theorem can be proved without the axiom of choice.)
However, it appears to be the case that the theorem is equivalent to the axiom of dependent choice. See, for instance, this answer. The statement in the book would be true in the sense that the full-strength axiom of choice is not required, but since this is an axiomatic set theory book, a statement like this seems like it would imply it's provable under ZF.
I'm wondering if this is just an oversight, or is it that they are using a weak version of the theorem that doesn't require any choice axiom? Their version of the theorem is given toward the end of section $0$ in the same chapter (p. 144):
Given any denumerable set $S$ of pure formulas (i.e. w/o constants), if there is a relational system in which all the formulas of $S$ are satisfiable, then there is a denumerable such relational system.
This version indeed does not require the axiom of choice. The more usual formulation that is equivalent to DC requires that the countable model you obtain be an elementary submodel of your original model. The version you quoted can be proved without choice as follows. First, if $S$ has a model, then it is consistent. Henkin's proof of the completeness theorem (which does not require choice if your language is countable) then produces a countable model of $S$.