Choice Function Exists for Well-Orderable Union of 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 the union $$\bigcup \mathbb S$$ be well-orderable.

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.

Corollary
The Well-Ordering Theorem, which states that every set is well-orderable, implies the truth of the Axiom of Choice.

Proof
Suppose $$T = \bigcup \mathbb S$$ is well-orderable.

Then we can create a well-ordering $$\preceq$$ on $$T$$ so as to make $$\left({T, \preceq}\right)$$ a well-ordered set.

From the definition of well-ordered set, every subset of $$T$$ is itself well-ordered.

From Subset of Union: General Result we have that $$\forall S \in \mathbb S: S \subseteq T$$.

So every $$S \in \mathbb S$$ is well-ordered and Choice Function Exists for Set of Well-Ordered Sets applies.

Proof of Corollary
Let $$\mathbb S$$ be any set of sets.

Assume the truth of the Well-Ordering Theorem.

That is, that every set is well-orderable.

Then, in particular, $$\bigcup \mathbb S$$ is well-orderable.

The result follows directly from the main theorem.