Set of Pairwise Disjoint Intervals is Countable

Theorem
Let $X$ be a subset of $\mathcal P \left({\R}\right)$ such that
 * $X$ is mutually disjoint:
 * $\forall A,B \in X: A \ne B \implies A \cap B = \varnothing$.
 * every element of $X$ contains an open interval:
 * $\forall A \in X: \exists x, y \in \R: x < y \land \left({x \,.\,.\, y}\right) \subseteq A$.

Then $X$ is countable.

Proof
by Between two Real Numbers exists Rational Number:
 * $\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \left({x \,.\,.\, y}\right) \subseteq A$.

By the Axiom of Choice define a mapping $f: X \to \Q$:
 * $\forall A \in X: f \left({A}\right) \in A$.

We will prove that $f$ is an injection by definition.

Let $A, B \in X$ such that
 * $f \left({A}\right) = f \left({B}\right)$.

By definition of $f$;
 * $f \left({A}\right) \in A$ and $f \left({B}\right) \in B$.

By definition of intrsection:
 * $f \left({A}\right) \in A \cap B$.

Then by definition of empty set:
 * $A \cap B \ne \varnothing$.

Thus by definition of mutually disjoint:
 * $A = B$.

This ends the proof of injection.

By Cardinality of Image of Injection:
 * $\left\vert{X}\right\vert = \left\vert{f \left({X}\right)}\right\vert$ because $X$ is subset of $X$.

By definition of image
 * $f \left({X}\right) \subseteq \Q$.

By Rational Numbers are Countably Infinite:
 * $\Q$ is countable.

Hence by Subset of Countable Set is Countable:
 * $f \left({X}\right)$ is countable.

Thus by Set is Countable if Cardinality equals Cardinality of Countable Set the result:
 * $X$ is countable.