Properties of axiom of choice Let us suppose that $\forall x \exists \alpha\exists f$ such that $f: \alpha\xrightarrow{}x$ where $f$ is bijection. Mark this property with $*$.
I have to show that this implies Axiom of Choice.
I have been given the proof in my lecture notes, but I don't get one part of it. 
Namely, the proof is given by:
Let the set $x$ consist of nonempty pairwise disjoint elements. Apply $*$ to $\cup x$.
Take an ordinal $\alpha$ and a function $f:\alpha\xrightarrow{}\cup x$. Define a choice set $z$ for $x$ by setting 
$z=${$f(y) | \exists u \in x (f(y) \in u \wedge \forall w<y$ $f(w) \notin u)$}.
So $z$ chooses for every $u \in x$ that $f(y) \in u$ with $y$ minimal.
And here the proof ends.
My problem is next,
I don't get how the above implies that for EVERY $u$, $z$ chooses if by definition of $z$ we only know that there exists such $u$. Also, how this implies that for every $u \in x$, $z \cap u$ is singleton (from the way how we define AC.)
By the way, we defined AC as the statement:
$\forall x(\emptyset \notin x\wedge \forall u,v \in x (u\neq v \implies u \cap v = \emptyset) \implies \exists z \forall u \in x \exists w$ $ u \cap z =${$w$}$)$
 A: The AC as you've defined it says that if $x$ is a family of non-empty ($0 \not \in x$) and pairwise-disjoint sets ($u \neq v \Rightarrow u \cap v = \emptyset$), then there exists a set $z$ such that $u \cap z$ is singleton for every $u \in x$. Now supposing the hypothesis has obtained, i.e. supposing $x$ is a family of pairwise-disjoint nonempty sets, then we need to show we've constructed $z$ with the property you've described. Let's define a bijection $g : x \to z$ by $$g(u) = f( \min \{ y \in \alpha : f(y) \in u \} ) .$$
We can check that $g(u) \in z$, where $z$ is as you described it earlier, by noting that $g(u) = f( \min \{ y \in \alpha : f(y) \in u \} ) \in u$, and if $w < \min \{ y \in \alpha : f(y) \in u \}$ then $f(w) \not \in u$.
Now we check that $u \cap z$ is exactly the singleton $\{ g(u) \}$. Evidently $g(u) \in u \cap z$, but assume for contradiction that there existed $a \in z$ such that $a \neq g(u)$ and $a \in u \cap z$. Then if we set $f(y') = g(u) , a = f(y'')$, we have two options: either $y' > y''$, or $y' < y''$. The former is excluded because this would contradict that $y' = \min \{ y \in \alpha : f(y) \in u \}$; as for the latter, this would contradict that $a \in u$, as we'd have that there exists $w < y''$ such that $f(w) \in u$, namely $w = y'$. Thus we have that $u \cap z = \{ g(u) \}$ and nothing else.
