Is AC necessary to prove that whenever $S$ is a set of non-empty sets, there exist a set $A$ which for each $s\in S$, $s\cap A\neq\varnothing$?

Question

Is this theorem need Axiom of Choice ? or is it equal to that?

Let $S$ be set of non-empty sets, there exist a set $A$ which for each $s\in S$, $s\cap A\neq\varnothing$

I think for making a set "$S$" such that satisfy $\forall s\in S, s\cap A\neq\varnothing$ we need choice function and we need AoC but apparently i was wrong. I don't know where i made a mistake.

Answer 1

No, the axiom of choice is not needed, since $A=\bigcup S$ works. The mistake is that knowing that $A\cap s\neq\varnothing$ is not enough to build a choice function: this intersection might contain many elements for every $s$, and then you need to choose one for every $s$ in order to get a choice function

Answer 2

Just take $A=\bigcup S$ by the axiom of union.