Let $T = \struct {\R, \tau}$ be the Sorgenfrey line. Let $\CC$ be an open cover for $\R$. Define $\VV = \set {U^\circ_\R: U \in \CC}$ where $U^\circ_\R$ denotes the interior of $U$ in the real number line $R = \struct {\R, \tau_d}$ with the usual (Euclidean) topology. By definition of interior: :$\forall U \in \CC: U^\circ_\R \subseteq \R$ By Union is Smallest Superset: :$W := \bigcup \VV \subseteq \R$  By Topological Subspace of Real Number Line 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: :$\forall A \in \VV: A \subseteq W$ By Intersection with Subset is Subset: :$\forall A \in \VV: A \cap W = A$ By definition of topological subspace: :$\forall A \in \VV: A$ is open in $R_W$ By definition :$\VV$ is open cover By definition of Lindelöf space: :there exists a countable subcover $\SS$ of $\VV$ By definition of $\VV$: :$\forall A \in \VV: \exists U \in \CC: A = U^\circ_\R$ By Axiom of Choice define a mapping $g: \VV \to \C C$: :$\forall A \in \VV: A = \map g A^\circ_\R$ Define $K = \map {g^\to} \SS$ where $\map {g^\to} \SS$ denotes the image of $\SS$ under $g$. Define $Y := \R \setminus \bigcup K$ By definition of cover: :$\R \subseteq \bigcup \CC$ By definition of subset: :$\forall x \in \R: x \in \bigcup \CC$ By definition of union: :$\forall x \in \R: \exists U \in \CC: x \in U$ By Axiom of Choice define a mapping $f: \R \to \CC$ such that :$\forall x \in \R: x \in \map f x$ Define $\BB := \set {\hointr x y: x, y \in \R}$ By definition of the Sorgenfrey line: :$\BB$ is a basis of $T$. By definition of $f$: :$\forall x \in \R: \map f x \in \CC$ By definition of open cover: :$\forall x \in \R: \map f x$ is open By definition of a basis: :$\forall x \in \R: \exists U_x \in \BB: x \in U_x \subseteq \map f x$ By definition of $\BB$: :$\forall x \in \R: \exists y, z \in \R: x \in \hointr y z \subseteq \map f x$ By definition of half-open real interval: :$\forall x \in \R: \exists z \in \R: x \in \hointr x z \subseteq \map f x$ By Axiom of Choice define a mapping $k: \R \to \BB$: :$\forall x \in \R: \exists z \in \R: x \in \map k x = \hointr x z \subseteq \map f x$ We will prove that :$(1): \quad \forall x, y \in Y: x \ne y \implies \paren x \cap \map k y = \O$ {{Mistake|For $\paren x$ should it really be $\map k x$?}} Let $x, y \in Y$ such that $x \ne y$. {{AimForCont}}: :$\paren x \cap \map k y \ne \O$ By definitions of empty set and intersection: :$\exists s: s \in \map k x \land s \in \map k y$ By definition of $k$: :$\exists z_1 \in \R: \map k x = \hointr x {z_1} \subseteq \map f x$ and :$\exists z_2 \in \R: \map k y = \hointr y {z_2}$ By Trichotomy Law for Real Numbers: :$x < y$ or $x > y$ {{WLOG}}, 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: :$y \in \openint x {z_1}$ By Open Real Interval is Open Set: :$\openint x {z_1}$ is topologically open in $R$ By definition of subset: :$\openint x {z_1} \subseteq \hointr x {z_1}$ By Subset Relation is Transitive: :$\openint x {z_1} \subseteq \map f x$ By Interior of Subset: :$\openint x {z_1}^\circ_\R \subseteq \map f x^\circ_\R$ By Interior of Open Set: :$\openint x {z_1}^\circ_\R \openint x z_1$ {{mistake|The relation is missing in the above -- subset or equals? Research needed}} By definition of subset: :$y \in \map f x^\circ_\R$ By definition of $\VV$: :$\map f x^\circ_\R \in \VV$ By definition of union: $y \in W$ By definition of interior: :$\forall A \in \SS: A \subseteq \map g A$ By Set Union Preserves Subsets: :$W \subseteq \bigcup \SS \subseteq \bigcup K$ By definition of subset: :$y \in \bigcup K$ This contradicts $y \in Y$ by definition of difference. Thus: :$\map k x \cap \map k y = \O$ By Set of Pairwise Disjoint Intervals is Countable :$\map {k^\to} Y$ is countable We will prove that :$k \restriction_Y$ is an injection Let $x, y \in Y$ such that :$\map {k \restriction_Y} x = \map {k \restriction_Y} Y$ {{AimForCont}}: :$x \ne y$ Then by $(1)$: :$\map k x \cap \map k y = \O$ By definition of restriction of mapping: :$\map {k \restriction_Y} x = \map k x$ and :$\map {k \restriction_Y} y = \map k y$ By definition of $k$: :$x \in \map k x$ So: :$x \in \map k y$ This contradicts: :$\map k x \cap k \map k y = \O$ Thus $x = y$. By Injection to Image is Bijection: :$k \restriction_Y: Y \to \map {k^\to} Y$ is a bijection By definitions of set equivalence and cardinality: :$\card Y = \card {\map {k^\to} Y}$ where $\card Y$ denotes the cardinality of $Y$. By Cardinality of Image of Set not greater than Cardinality of Set: :$\card K \le \card \SS$ and $\card {\map {f^\to} Y} \le \card Y$ By Countable iff Cardinality not greater than Aleph Zero: :$\card \SS \le \aleph_0$ and $\card Y \le \aleph_0$ Then: :$\card K \le \aleph_0$ and $\card {\map {f^\to} Y} \le \aleph_0$ By Countable iff Cardinality not greater than Aleph Zero: :$K$ is countable and $\map {f^\to} Y$ is countable Thus by Countable Union of Countable Sets is Countable: :$\GG := K \cup \map {f^\to} Y$ is countable By definition of image of set: :$K \subseteq \CC$ and $\map {f^\to} Y \subseteq \CC$ thus by corollary of Set Union Preserves Subsets: :$\GG \subseteq \CC$ It remains to prove that :$\GG$ is cover for $\R$ Let $x \in \R$. By Union Distributes over Union: Sets of Sets: :$\bigcup \GG = \paren {\bigcup K} \cup \bigcup \map {f^\to} Y$ {{AimForCont}}: :$ x \notin \bigcup \GG$ By definition of union: :$x \notin \bigcup K$ and $x \notin \bigcup \map {f^\to} Y$ By definition of difference: :$x in Y$ By definition of image of set: :$\map f x \in \map {f^\to} Y$ By definition of $f$: :$x \in \map f x$ By definition of union: :$x \in \bigcup \map {f^\to} Y$ This contradicts $x \notin \bigcup \map {f^\to} Y$. Thus the result by Proof by Contradiction. {{qed}}