Why we don't need Axiom of Choice to prove the following statement
Let $S_{\alpha}, \alpha \in A$ be a family of disjoint nonempty sets, and consider $P = \bigcup_{\alpha \in A} S_{\alpha}$. If $|A|$ is finite then there exists $Q \subset P$ such that for each $\alpha \in A$, we have $|Q \cap S_{\alpha}| = 1$
with this as the proof (taken from https://math.stackexchange.com/a/29383/)
Since each of the $S_\alpha$'s are nonempty, then by definition for each $\alpha$ there exits $b_{\alpha} \in S_{\alpha}$. So $Q = \{b_{\alpha} | \alpha \in A \}$ works.
But apparently we do need Axiom of Choice to prove the exact same hypothesis with just the hypothesis ``$|A|$ is finite" removed.
Can someone provide some intuition on why the proof won't work for infinite $A$ ?
You're choosing the $b_\alpha$. If you can specify a way to single out a unique $b_\alpha$ for each $\alpha$ (including if $|A|$ is finite, and you, say, list the $b_\alpha$ explicitly), the axiom schema of replacement can tell you that $\{b_\alpha\mid \alpha\in A\}$ is a set. If not, then none of the ZF axioms let you say that $\{b_\alpha\mid \alpha\in A\}$ (whatever it now is) is a set. That's exactly what the Axiom of Choice does.
So without Chioce, and with infinite $|A|$, in general a collection of the form $\{b_\alpha\mid \alpha\in A\}$ will not be a set. With Choice, you know that there is at least one such set, although it doesn't say in any way how the $b_\alpha$ in such a set were chosen.