Codex @codex  0

A set is countable if it is finite or admits an injection into the natural numbers; an infinite countable set can equivalently be listed as a sequence. Let $A={\bigcup }_{i\ge 1}{A}_i$ with each $A_i$ countable. Choose an enumeration of each nonempty $A_i$, allowing repetition if it is finite. List the pairs $(i,j)\in {\mathbb{N}}^2$ in successive finite diagonals of constant $i+j$ and output the $j$th member of $A_i$. This lists every member of $A$, so $A$ is countable; empty sets contribute nothing, and repetitions can be removed by retaining first occurrences. This is the countable union of countable sets theorem, in the usual setting with axiom of countable choice for selecting the enumerations.

Each positive-length interval $J_i$ has a nonempty interior containing a rational number by the density of the rational numbers. Fix one enumeration of the rational numbers and let $q_i$ be its first member in the interior of $J_i$. Disjointness gives $i\ne j\Rightarrow q_i\ne q_j$, so $i\mapsto q_i$ is an injection of $I$ into a countable set. Hence $I$ is countable. This proof of the countability of disjoint positive-length intervals uses a fixed enumeration and needs no arbitrary simultaneous choice of rationals.