You are asking the following:
Does $\sf ZF$ proves that "Every infinite set has a countably infinite subset"? If not, then does $\sf ZF$ with "Every infinite set has a countably infinite subset" prove the axiom of choice?
The answer to both is negative.
Both proofs are non-trivial, and require substantial knowledge of independence proofs and models of $\sf ZF$. But just minor bits of history:
Fraenkel proved that if it consistent that the answer of the first question is false, and the axiom of choice fails, but only if we weaken set theory to allow objects which are not sets (called atoms, or urelements).
Cohen proved this in $\sf ZF$ some 40 years afterwards in his original papers about forcing. He took the ideas from Fraenkel's work (after it was polished and corrected by Mostowski and Specker in the meantime) and applied them together with his new method of forcing.
You can find both these in the book "The Axiom of Choice" by Jech.
In the other direction, Jech's historical notes from his "Axiom of Choice" book say that Pincus proved this in 1969 (see p.132 in the book). I also know that Monro also proved this in 1975 directly using forcing and relative constructibility arguments.
All proofs mentioned here are not trivial at all, and without further knowledge about forcing, relative constructibility, permutation models and symmetric models, it's hard to explain how these proofs go. But if you want to study these topics, the book mentioned by Jech is a good start, as well his "Set Theory" book, and Halbeisen's book "Combinatorial Set Theory".