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

Define $Y := \R \setminus \bigcup K$

We will prove that
 * every element of $Y$ is a local minimum in $Y$

Let $y \in Y$.

Aiming for a contradiction suppose that
 * $y$ is not a local minimum in $Y$

By definition of local minimum:
 * $\lnot \exists z \in \R: z < y \land \left({z\,.\,.\,y}\right) \cap Y = \varnothing$

Then:
 * $\exists z \in \R: z < y \land \left({z\,.\,.\,y}\right) \subseteq Y$

By Set of Local Minimum is Countable:
 * $Y$ is countable

By Cardinality of Image of Set not greater than Cardinality of Set:
 * $\left\vert{K}\right\vert \le \left\vert{\Q}\right\vert$ and $\left\vert{f^\to\left({Y}\right)}\right\vert \le \left\vert{Y}\right\vert$

where $\left\vert{K}\right\vert$ denotes the cardinality of $K$.

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

By Countable iff Cardinality not greater than Aleph Zero:
 * $\left\vert{\Q}\right\vert \le \aleph_0$ and $\left\vert{Y}\right\vert \le \aleph_0$

Then:
 * $\left\vert{K}\right\vert \le \aleph_0$ and $\left\vert{f^\to\left({Y}\right)}\right\vert \le \aleph_0$

By Countable iff Cardinality not greater than Aleph Zero:
 * $K$ is countable and $f^\to\left({Y}\right)$ is countable

Thus by Countable Union of Countable Sets is Countable:
 * $\mathcal G := K \cup f^\to\left({Y}\right)$ is countable

By definition of image of set:
 * $K \subseteq \mathcal C$ and $f^\to\left({Y}\right) \subseteq \mathcal C$

thus by corollary of Set Union Preserves Subset:
 * $\mathcal G \subseteq \mathcal C$

It remains to prove that
 * $\mathcal G$ is cover for $\R$

Let $x \in \R$.

By Union Distributes over Union/Sets of Sets:
 * $\bigcup \mathcal G = \left({\bigcup K}\right) \cup \bigcup f^\to\left({Y}\right)$

Aiming for a contradiction suppose that
 * $ x \notin \bigcup \mathcal G$

By definition of union:
 * $x \notin \bigcup K$ and $x \notin \bigcup f^\to\left({Y}\right)$

By definition of difference:
 * $x in Y$

By definition of image of set:
 * $f\left({x}\right) \in f^\to\left({Y}\right)$

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

By definition of union:
 * $x \in \bigcup f^\to\left({Y}\right)$

This contradicts $x \notin \bigcup f^\to\left({Y}\right)$

Thus the result by Proof by Contradiction.