Union of Countable Sets of Sets/Proof 3

Proof
Since $\AA$ and $\BB$ are countable, $\AA \times \BB$ is countable by Cartesian Product of Countable Sets is Countable.

Thus by Surjection from Natural Numbers iff Countable there is a surjection $f: \N \to \AA \times \BB$.

Let $\CC = \set {A \cup B: A \in \AA, B \in \BB}$.

Let $g: \AA \times \BB \to \CC$ be defined by letting $\map g {A, B} = A \cup B$.

$g$ is a surjection:

Let $C \in \CC$.

Then there exist $A \in \AA$ and $B \in \BB$ such that $C = A \cup B$.

By the definition of Cartesian product:
 * $\tuple {A, B} \in \AA \times \BB$

Then:
 * $\map g {A, B} = A \cup B = C$

It follows that $g$ is a surjection.

By Composite of Surjections is Surjection, $g \circ f: \N \to \CC$ is surjective.

Hence $\CC$ is countable.