3 Statements of axiom of choice are equivalent

$3\Rightarrow 2$: Suppose $X$ is a collection of nonempty sets, then $f:P(\bigcup X)\setminus\{\varnothing\}\to\bigcup X$ exists which chooses from every nonempty subset of $\bigcup X$. If we restrict $f$ to the set $X$ then we have a choice function.

$2\Rightarrow 3$: Trivial, if every non-empty collection of non-empty sets has a choice functions, given a non-empty $X$ we have that $P(X)\setminus\{\varnothing\}$ is a non-empty collection of non-empty sets, thus has a choice function.

$2\Rightarrow 1$: Trivial, if every collection has a choice function then every pairwise disjoint collection has a choice function $\{f(x)\mid x\in X\}$ is a choice set.

$1\Rightarrow 2$: Given $X$ a non-empty collection of non-empty sets, let $\{\{x\}\times x\mid x\in X\}$ be a collection of now pairwise disjoint sets. The cut set is a choice function. (You may want to show why they are pairwise disjoint, this is because if $x\neq y$ then $(\{x\}\times x)\cap(\{y\}\times y)=\varnothing$; also you might want to show why the choice set is a function, but it only meets $\{x\}\times x$ at one point... so functionality holds and it is indeed a choice function.)

I'll show you how to prove 3) implies 1), the rest should be relatively easy. So let $X$ be a nonempty set. For every nonempty subset $Y$ of $X$, let $Y^*=Y\times\{Y\}$. Then the family $\{Y^*:X\supseteq Y\neq\emptyset\}$ consists of disjoint nonempty sets, so there exists a choice set $C$. Let $f^*$ be the function that maps each nonempty set $Y\subseteq X$ to the unique element of $C$ that lies in $Y^*$. This isn't yet the function we want to construct. So let $\pi$ be the function that maps each element in $\{Y\times\{Y\}:X\supseteq Y\neq\emptyset\}$ (a ordered pair) to its first coordinate. Then letting $f=\pi\circ f^*$ gives you the needed function, as you can easily verify.