Sorgenfrey Line is Lindelöf

Theorem
Sorgenfrey line is Lindelöf.

Proof
Let $T = \left({\R, \tau}\right)$ be the Sorgenfrey line.

Let $\mathcal C$ be an open cover for $\R$.

By definition of cover:
 * $\R \subseteq \bigcup \mathcal C$

By definition of subset:
 * $\forall x \in \R: x \in \bigcup \mathcal C$

By definition of union:
 * $\forall x \in \R: \exists U \in \mathcal C: x \in U$

By Axiom of Choice define a mapping $f: \R \to \mathcal C$ such that
 * $\forall x \in \R: x \in f\left({x}\right)$

Define $K = f^\to\left({\Q}\right)$

where $f^\to\left({\Q}\right)$ denotes the image of $\Q$ under $f$.

Define $\mathcal B := \left\{{\left[{x\,.\,.\,y}\right): x, y \in \R}\right\}$

By definition of the Sorgenfrey line:
 * $\mathcal B$ is a basis of $T$.

By definition of $f$:
 * $\forall x \in \R: f\left({x}\right) \in \mathcal C$

By definition of open cover:
 * $\forall x \in \R: f\left({x}\right)$ is open

By definition of a basis:
 * $\forall x \in \R: \exists U_x \in \mathcal B: x \in U_x \subseteq f\left({x}\right)$

By definition of $\mathcal B$:
 * $\forall x \in \R: \exists y, z \in \R: x \in \left[{y\,.\,.\,z}\right) \subseteq f\left({x}\right)$