prove Cartesian product of countable non empty sets is not empty without Zorn's Lemma I saw this question in the book A Taste of Topology by Volker Runde: 
Let $(S_n)^{\infty}_{n=1}$ be a sequence of nonempty sets. Show without invoking Zorn's lemma that $\prod^{\infty}_{n=1}S_n$ is not empty. 
I'm puzzled as to how this can be done without Zorn's Lemma/Axiom of Choice.
 A: As the comments have shown, the statement is equivalent to the axiom of countable choice which is independent from ZF (see for example https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory).
However, frequently the axiom of countable choice is confused with the principle of mathematical induction. I do not claim that Runde does it, but it is not impossible.
You will find many proofs in the literature where something is "constructed inductively". An example is the Bolzano-Weierstrass theorem (every bounded sequence of real numbers has a convergent subsequence). Here one makes infinitely many choices to obtain the desired subsequence, but usually this is not acknowledged as the use of the axiom of choice (in this case more precisely the axiom of dependent choice).
If you are interested in that subject see for instance
http://emis.ams.org/journals/CMUC/pdf/cmuc9703/herrli.pdf
http://www.mat.uc.pt/~ggutc/teses/teseingles.pdf
https://dml.cz/bitstream/handle/10338.dmlcz/105069/CommentatMathUnivCarol_007-1966-3_11.pdf
Keremedis, Kyriakos. "Disasters in topology without the axiom of choice." Archive for Mathematical Logic 40.8 (2001): 569-580
