I would like to prove the following:
Let $M$ be a non-finitely generated right module over a ring $R$, and presume axiom of choice is true. Then there exists a subset of $M$, such that it generates $M$, and it's minimal as far as subsets that generate $M$ go.
The discussion that follows in the comments in the link above didn't clear up much, as it seemed too advanced for me. Also I'm not too interested in whether full choice is needed etc., I'm fine with presuming full choice.
Here's what I've done so far. The idea is we use Zorn's lemma on subsets of linearly independent elements. We get a maximal element, a subset $A$ of independent elements. If we could show that this subset generates the whole module, we would be done, because any proper subset would not generate the whole module (if that were not true, we would get a contradiction with independence of $A$).
However, I don't know how to show that $A$ actually generates the whole module. Presuming it's not, if we take some element $c$ that's not in $A$, it doesn't seem like I can simply add $c$ to $A$ (in hopes of getting a contradiction with maximality of $A$), because I don't really know that $cR \cap A=0$.