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 $$f: \mathbb S \to \bigcup \mathbb S$$ defined as:
 * $$\forall S \in \mathbb S: \exists x \in S: f \left({S}\right) = x$$

Thus, if every member of $$\mathbb S$$ is a well-ordered, then we can create a choice function $$f$$ defined as:
 * $$\forall S \in \mathbb S: f \left({S}\right) = \inf \left({S}\right)$$

True, we may be making infinitely many choices, but we have a rule for doing so.