Countable union of countable sets is countable

Statement

Assume the axiom of choice. Let $I$ be a countable set and let $(A_i)_{i \in I}$ be a family of sets such that each $A_i$ is countable. Then the union $\bigcup_{i \in I} A_i$ is countable.

Remarks

Proof: choose for each $i$ a surjection from $\mathbb{N}$ onto $A_i$, when $A_i$ is nonempty. Together they give a surjection from a subset of $\mathbb{N} \times \mathbb{N}$ onto the union, and $\mathbb{N} \times \mathbb{N}$ is countable. The proof uses the axiom of countable choice to choose the surjections. Without any choice, the statement cannot be proved in ZF. When the enumerations of the $A_i$ are given, no choice is needed.

Explanations

L1 human

List each set $A_i$ as a row of a grid. Then list the grid diagonal by diagonal. Each element of the union appears in the list.