Let $\CC$ be a open cover for $W$. Define $Q := \set {\openint a b: a, b \in \Q}$ Define a mapping $h: Q \to \Q \times \Q$: :$\forall \openint a b \in Q: \map h {\openint a b} = \tuple {a, b}$ It is easy to see by definition that :$h$ is an injection. By Injection iff Cardinal Inequality: :$\card Q \le \card {\Q \times \Q}$ where $\card Q$ 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: :$\card \Q = \card \N$ By Aleph Zero equals Cardinality of Naturals: :$\card \Q = \aleph_0$ By Cardinal Product Equal to Maximum: :$\card {\Q \times \Q} = \max \set {\aleph_0, \aleph_0} = \aleph_0$ By Countable iff Cardinality not greater than Aleph Zero: :$Q$ is countable. By definition of cover: :$W \subseteq \bigcup \CC$ By definition of imion: :$\forall x \in W: \exists U \in \CC: x \in U$ By Axiom of Choice define a mapping $f: W \to \CC$: :$\forall x \in W: x \in \map f x$ We will prove that :$\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq \map f x$ Let $x \in W$. By definition of open cover: :$\map f x$ is open in $R_W$. By definition of topological subspace: :there exists U a subset of $\R$ such that ::$U$ is open in $R$ and $U \cap W = \map f x$ By definition of $f$: :$x \in \map f x$ By definition of open set in metric space: :$\exists r > 0: \map {B_r} x \subseteq U$ :$\openint {x - r} {x + r} \subseteq U$ 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 $Q$: :$\openint q p \in Q$ Thus by definition of open real interval: :$x \in \openint q p \subseteq \openint {x - r} {x + r}$ By Subset Relation is Transitive: :$\openint q p \subseteq U$ Thus by Set Intersection Preserves Subsets/Corollary: :$\openint q p \cap W \subseteq \map f x$ By Axiom of Choice define a mapping $f_1: W \to Q$: :$\forall x \in W: x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$ By definitions of image of set and image of mapping: :$\forall A \in \Img {f_1}: \exists x \in W: x \in f_1^{-1} \sqbrk {\set A}$ By Axiom of Choice define a mapping $c: \Img {f_1} \to W$: :$\forall A \in \Img {f_1}: \map c A \in f_1^{-1} \sqbrk {\set A}$ Define a mapping $g: \Img {f_1} \to \CC$: :$g := f \circ c$ Define $\GG = \Img g$. Thus $\GG \subseteq \CC$ By definition of image of mapping: :$\Img {f_1} \subseteq Q$ By Subset of Countable Set is Countable; :$\Img {f_1}$ is countable. By Surjection iff Cardinal Inequality: :$\size {\Img g} \le \size {\Img {f_1} }$ Thus by Countable iff Cardinality not greater than Aleph Zero: :$\GG$ is countable. It remains to prove that :$\GG$ is a cover for $W$. Let $x \in W$. By definition of $f_1$: :$ x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$ By definition of image of mapping: :$\map {f_1} x \in \Img {f_1}$ Then by definition of $c$: :$y := \map c {\map {f_1} x} \in f_1^{-1} \sqbrk {\set {\map {f_1} x} }$ By definition of $f_1$: :$y \in \map {f_1} y \land \map {f_1} y \cap W \subseteq \map f y$ By definition of image of set: :$\map {f_1} y = \map {f_1} x$ Then by definitions of subset and intersection: :$x \in \map f y$ By definition of composition of mappings: :$\map f y = \map g {\map {f_1} x}$ By definition of image of mapping: :$\map g {\map {f_1} x} \in \GG$ Thus by definition of union: :$x \in \bigcup \GG$ Thus the result. {{qed}}