I want to prove that the following variants of the Axiom of Choice are equivalent:
AC1. Let $X$ and $Y$ be sets and let $F:X\rightarrow\wp\left(Y\right)-\left\{ \emptyset\right\}$ be a function, where $\wp\left(Y\right)$ is the power set of $Y$. Then there exists a function $f:X\rightarrow Y$ such that $f\left(x\right)\in F\left(x\right)$ for all $x\in X$.
and
AC2. If $X$ is a non-empty set, then there exists a function $f:X\rightarrow\bigcup X$ such that $f\left(A\right)\in A$ for all $A\in X$.
I have proven that AC2 implies AC1 but I am having trouble proving that AC1 implies AC2. I know that I have to chose an appropriate $Y$ and $F$, but I don't know which ones.
Take $Y$ to be $\bigcup X$, and $F$ to be the identity function. Then apply AC1 to deduce AC2.