Set of Pairwise Disjoint Intervals is Countable

Theorem
Let $X$ be a subset of $\powerset \R$ such that:
 * $(1): \quad X$ is pairwise disjoint:
 * $\forall A, B \in X: A \ne B \implies A \cap B = \O$.
 * $(2): \quad$ every element of $X$ contains an open interval:
 * $\forall A \in X: \exists x, y \in \R: x < y \land \openint x y \subseteq A$.

Then $X$ is countable.

Proof
By Between two Real Numbers exists Rational Number:
 * $\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \openint x y \subseteq A$

By the Axiom of Choice define a mapping $f: X \to \Q$:
 * $\forall A \in X: \map f A \in A$

First it needs to be shown that $f$ is an injection by definition.

Let $A, B \in X$ such that:
 * $\map f A = \map f B$

By definition of $f$:
 * $\map f A \in A$ and $\map f B \in B$

By definition of intersection:
 * $\map f A \in A \cap B$

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

Thus by definition of pairwise disjoint:
 * $A = B$

Hence $f$ is an injection.

By Set is Subset of Itself, $X$ is a subset of $X$.

Thus by Cardinality of Image of Injection:
 * $\card X = \card {f^\to \sqbrk X}$

By definition of image:
 * $f^\to \sqbrk X \subseteq \Q$

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

Hence by Subset of Countable Set is Countable:
 * $f^\to \sqbrk X$ is countable.

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