I'm reading Zorich'es "Mathematical Analysis I", Ed 4, 2004, and wonder if this is a trifle misinterpretation of "Axiom of Choice". Ch 1.4 "Supplementary Material" says:
8°. (A x i o m o f c h o i c e) For any family of nonempty sets there exists a set $C$ such that for each set $X$ in the family $X \cap C$ consists of exactly one element.
I suppose for nonempty set family $\{A,B,S\}$, if $A, B \subset S$ and $A\cap B=\emptyset$, $S\cap C$ will have at least 2 elements?
Of course this is a trifle one and I'm not sure if it's from Zorich or translation.
Yes, this is a mistake.
The formulation would be that for every family of non-empty, pairwise disjoint sets there is a set $C$, that for each $X$ in the family, $X\cap C$ has exactly one element.
To see that there is a problem, pick any set of at least two elements $A$ and consider $\mathcal P(A)\setminus\{\varnothing\}$ as your family of non-empty sets. It is not hard to see that $C$ would have to be $A$ itself, otherwise it wouldn't meet the singletons, but then $A\cap A=A$ has more than one elements.