Axiom of Choice equivalents

So apparently for all pairwise disjoint sets S not containing the empty set

$\displaystyle \exists f [\exists Y (f:S \rightarrow Y) and \forall x (x \in S \implies f(x) \in x)] \iff \exists t [\forall z z \in S \implies \exists! w (w \in z \cap t)]$

I had no problems proving the first version from the second, and assuming the first one and taking f(S) (image of the whole set S under f) for t, then obviously for all z in S f(z) is in z and f(S). The reason I'm posting is because I can't for the life of me prove that it's the only element of z intersection f(S)!