Topological Subspace of Real Number Line is Lindelöf

Theorem
Let $R = \left({\R, \tau_d}\right)$ be the Real number topological space.

Let $W$ be a non-empty subset of $\R$.

Then $R_W$ is Lindelöf

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

Proof
Let $\mathcal C$ be a open cover for $W$.

Define $Q := \left\{{\left({a\,.\,.\,b}\right): a, b \in \Q}\right\}$

Define a mapping $h:Q \to \Q\times\Q$:
 * $\forall \left({a\,.\,.\,b}\right) \in Q:h \left({\left({a\,.\,.\,b}\right)}\right) = \left({a, b}\right)$

It is easy to see by definition that
 * $h$ is an injection.

By Injection iff Cardinal Inequality:
 * $\left\vert{Q}\right\vert \le \left\vert{\Q \times \Q}\right\vert$

where $\left\vert{Q}\right\vert$ deontes the cardinality of $Q$.

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

By definition of countably infinite:
 * there exists a bijection $\Q \to \N$

By definitions of set equality and cardinality:
 * $\left\vert{\Q}\right\vert = \left\vert{\N}\right\vert$

By Aleph Zero equals Cardinality of Naturals:
 * $\left\vert{\Q}\right\vert = \aleph_0$

By Cardinal Product Equal to Maximum:
 * $\left\vert{\Q \times \Q}\right\vert = \max\left({\aleph_0, \aleph_0}\right) = \aleph_0$

By Countable iff Cardinality not greater than Aleph Zero:
 * $Q$ is countable.

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

By definition of imion:
 * $\forall x \in W: \exists U \in \mathcal C: x \in U$

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

We will prove that
 * $\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq f \left({x}\right)$

Let $x \in W$.

By definition of open cover:
 * $f \left({x}\right)$ os open in $R_W$.

By definition of topological subspace:
 * there exists U a subset of $\R$ such that
 * $U$ is open in $R$ and $U \cap W = f \left({x}\right)$

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

By definition of open set in metric space:
 * $\exists r > 0: B_r\left({x}\right) \subseteq U$


 * $\left({x-r\,.\,.\,x+r}\right) \subseteq U$

By Between two Real Numbers exists Rational Number:
 * $\exists q \in \Q: x-r < q < x$ and $\exists p \in \Q: x < p < x+r$

By definition of $Q$:
 * $\left({q\,.\,.\,p}\right) \in Q$

Thus by definition of open real interval:
 * $x \in \left({q\,.\,.\,p}\right) \subseteq \left({x-r\,.\,.\,x+r}\right)$

By Subset Relation is Transitive:
 * $\left({q\,.\,.\,p}\right) \subseteq U$

Thus by Set Intersection Preserves Subsets/Corollary:
 * $\left({q\,.\,.\,p}\right) \cap W \subseteq f \left({x}\right)$

By Axiom of Choice define a mapping $f_1:W \to Q$:
 * $\forall x \in W: x \in f_1\left({x}\right) \land f_1\left({x}\right) \cap W \subseteq f\left({x}\right)$

By definitions of image of set and image of mapping:
 * $\forall A \in \operatorname{Im}\left({f_1}\right): \exists x \in W: x \in f_1^{-1}\left[ {\left\{{A}\right\} }\right]$

By Axiom of Choice define a mapping $c: \operatorname{Im}\left({f_1}\right) \to W$:
 * $\forall A \in \operatorname{Im}\left({f_1}\right): c\left({A}\right) \in f_1^{-1}\left[ {\left\{{A}\right\} }\right]$

Define a mapping $g: \operatorname{Im}\left({f_1}\right) \to \mathcal C$:
 * $ g := f \circ c$

Define $\mathcal G = \operatorname{Im}\left({g}\right)$.

Thus $\mathcal G \subseteq \mathcal C$

By definition of image of mapping:
 * $\operatorname{Im} \left({f_1}\right) \subseteq Q$

By Subset of Countable Set is Countable;
 * $\operatorname{Im} \left({f_1}\right)$ is countable.

By Surjection iff Cardinal Inequality:
 * $\left\vert{\operatorname{Im}\left({g}\right)}\right\vert \le \left\vert{\operatorname{Im}\left({f_1}\right)}\right\vert$

Thus by Countable iff Cardinality not greater than Aleph Zero:
 * $\mathcal G$ is countable.

It remains to prove that
 * $\mathcal G$ is a cover for $W$.

Let $x \in W$.

By definition of $f_1$:
 * $ x \in f_1\left({x}\right) \land f_1\left({x}\right) \cap W \subseteq f\left({x}\right)$

By definition of image of mapping:
 * $f_1\left({x}\right) \in \operatorname{Im}\left({f_1}\right)$

Then by definition of $c$:
 * $y := c \left({f_1\left({x}\right)}\right) \in f_1^{-1}\left[ {\left\{ {f_1\left({x}\right)}\right\} }\right]$

By definition of $f_1$:
 * $ y \in f_1\left({y}\right) \land f_1\left({y}\right) \cap W \subseteq f\left({y}\right)$

By definition:Image/Relation/Subset|image os set]]:
 * $f_1\left({y}\right) =f_1\left({x}\right)$

Then by definitions of subset and intersection:
 * $x \in f \left({y}\right)$

By definition of composition of mappings:
 * $f \left({y}\right) = g \left({f_1\left({x}\right)}\right)$

By definition of image of mapping:
 * $g \left({f_1\left({x}\right)}\right) \in \mathcal G$

Thus by definition of union:
 * $x \in \bigcup \mathcal G$

Thus the result.