Choice Function Exists for Set of Well-Ordered Sets

Theorem

Let $\mathbb S$ be a set of sets such that:

$\forall S \in \mathbb S: S \ne \varnothing$

that is, none of the sets in $\mathbb S$ may be empty.

Let every element of $\mathbb S$ be well-ordered.

Then there exists a choice function $f: \mathbb S \to \bigcup \mathbb S$ satisfying:

$\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$

Proof

Every member of $\mathbb S$ is a well-ordered set.

Thus, for $S \in \mathbb S$, there is a minimal element $s$ for $S$ (with respect to the ordering of $S$).

By Well-Ordering Minimal Elements are Unique, $s$ is unique.

Therefore, we can define $f$ by:

$\forall S \in \mathbb S: f \left({S}\right) = s$

$\blacksquare$