I am trying to prove the converse (underlined in red) by defining a choice function $f:\Bbb{F}\rightarrow \cup_{\alpha\in I}X_\alpha$ to be $f(X_\alpha)=x_\alpha$ such that $Y\cap X_\alpha=x_\alpha$ where $x_\alpha$ is a unique intersection between $Y$ and $X_\alpha$. Also, $\Bbb{F}=\{X_\alpha: \alpha\in I\}$. I do not understand the problem where the Axiom of Choice does not assume the sets $X_\alpha$ being disjoint. In a same post here Exercise on the Axiom of Choice, Prof. Brian Scott said that if the family of $X_\alpha$ is not pairwaise disjoint then there's no way we can get the $Y$ we want to use to define $f$. Why?
