Sorgenfrey Line is Lindelöf

Theorem
The Sorgenfrey line is Lindelöf.

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

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

Define $\mathcal V = \left\{{U^\circ_\R: U \in \mathcal C}\right\}$

where $U^\circ_\R$ denotes the interior in $\R$ Euclidean topology $R = \left({\R,\tau_d}\right)$

By definition of interior:
 * $\forall U \in \mathcal C: U^\circ_\R \subseteq \R$

By Union is Smallest Superset:
 * $W := \bigcup \mathcal V \subseteq \R$

By Topological Subspace of Real Number Space is Lindelöf:
 * $R_W$ is Lindelöf

where $R_W$ denotes topological subspace of $R$ on $W$.

By Set is Subset of Union/Set of Sets:
 * $\forall A \in \mathcal V: A \subseteq W$

By Intersection with Subset is Subset:
 * $\forall A \in \mathcal V: A \cap W = A$

By definition of topological subspace:
 * $\forall A \in \mathcal V: A$ is open in $R_W$

By definition
 * $\mathcal V$ is open cover

By definition of Lindelöf space:
 * there exists countable subcover $\mathcal S$ of $\mathcal V$

By definition of $\mathcal V$:
 * $\forall A \in \mathcal V: \exists U \in \mathcal C: A = U^\circ_\R$

By Axiom of Choice define a mapping $g:\mathcal V \to \mathcal C$:
 * $\forall A \in \mathcal V: A = g\left({A}\right)^\circ_\R$

Define $K = g^\to\left({\mathcal S}\right)$

where $g^\to\left({\mathcal S}\right)$ denotes the image of $\mathcal S$ under $g$.

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

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

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

By definition of half-open real interval:
 * $\forall x \in \R: \exists z \in \R: x \in \left[{x\,.\,.\,z}\right) \subseteq f\left({x}\right)$

By Axiom of Choice define a mapping $k: \R \to \mathcal B$:
 * $\forall x \in \R: \exists z \in \R: x \in k\left({x}\right) = \left[{x\,.\,.\,z}\right) \subseteq f\left({x}\right)$

We will prove that
 * $(1): \quad \forall x, y \in Y: x \ne y \implies \left({x}\right) \cap k\left({y}\right) = \varnothing$

Let $x, y \in Y$ such that $x \ne y$.

Aiming for a contradiction suppose
 * $\left({x}\right) \cap k\left({y}\right) \ne \varnothing$

By definitions of empty set and intersection:
 * $\exists s: s \in k \left({x}\right) \land s \in k \left({y}\right)$

By definition of $k$:
 * $\exists z_1 \in \R: k \left({x}\right) = \left[{x\,.\,.\,z_1}\right) \subseteq f \left({x}\right)$

and
 * $\exists z_2 \in \R: k \left({y}\right) = \left[{y\,.\,.\,z_2}\right)$

By Trichotomy Law for Real Numbers:
 * $x < y$ or $x > y$

, suppose $x < y$

By definition of half-open real interval:
 * $x \le s < z_1$ and $y \le s < z_2$

Then: $y < z_1$

By definition of open real interval:
 * $y \in \left({x\,.\,.\,z_1}\right)$

By Open Real Interval is Open Set:
 * $\left({x\,.\,.\,z_1}\right)$ is topologically open in $R$

By definition of subset:
 * $\left({x\,.\,.\,z_1}\right) \subseteq \left[{x\,.\,.\,z_1}\right)$

By Subset Relation is Transitive:
 * $\left({x\,.\,.\,z_1}\right) \subseteq f \left({x}\right)$

By Interior of Subset:
 * $\left({x\,.\,.\,z_1}\right)^\circ_\R \subseteq f \left({x}\right)^\circ_\R$

By Interior of Open Set:
 * $\left({x\,.\,.\,z_1}\right)^\circ_\R \left({x\,.\,.\,z_1}\right)$

By definition of subset:
 * $y \in f \left({x}\right)^\circ_\R$

By definition of $\mathcal V$:
 * $f \left({x}\right)^\circ_\R \in \mathcal V$

By definition of union: $y \in W$

By definition of interior:
 * $\forall A \in \mathcal S: A \subseteq g \left({A}\right)$

By Set Union Preserves Subsets:
 * $W \subseteq \bigcup \mathcal S \subseteq \bigcup K$

By definition of subset:
 * $y \in \bigcup K$

This contradits $y \in Y$ by definition of difference.

Thus $k \left({x}\right) \cap k \left({y}\right) = \varnothing$

By Set of Pairwise Disjoint Intervals is Countable
 * $k^\to\left({Y}\right)$ is countable

We will prove that
 * $k \restriction_Y$ is an injection

Let $x, y \in Y$ such that
 * $k\restriction_Y \left({x}\right) = k\restriction_Y \left({y}\right)$

Aiming for a contradiction suppose that
 * $x \ne y$

Then by $(1)$:
 * $k \left({x}\right) \cap k \left({y}\right) = \varnothing$

By definition of restriction of mapping:
 * $k \restriction_Y \left({x}\right) = k\left({x}\right)$

and
 * $k \restriction_Y \left({y}\right) = k\left({y}\right)$

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

So:
 * $x \in k\left({y}\right)$

This contradicts:
 * $k \left({x}\right) \cap k \left({y}\right) = \varnothing$

Thus $x = y$.

By Restricting Injection to Bijection:
 * $k\restriction_Y:Y \to k^\to\left({Y}\right)$ is a bijection

By definitions of set equivalence and cardinality:
 * $\left\vert{Y}\right\vert = \left\vert{k^\to\left({Y}\right)}\right\vert$

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

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

By Countable iff Cardinality not greater than Aleph Zero:
 * $\left\vert{\mathcal S}\right\vert \le \aleph_0$ 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 Subsets:
 * $\mathcal G \subseteq \mathcal C$

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.