Proof of Euclidean Space is a Separable Metric Space

Each result cited below is universally quantified over the data in its own statement, and is used here for the dimension $n$ fixed in the statement. Write $\iota:\mathbb{N}\to\mathbb{R}$ for the canonical map of $\mathbb{R}$, as in The Archimedean Property of the Real Numbers, and let $\mathbb{Q}$ be the set of rational numbers.

1. Finite nets in the closed balls. Let $m\in\mathbb{N}$ and $s\in\mathbb{Q}$. If $0<\iota(m)$ and $0<s$, then the closed ball $\bar{B}(0,\iota(m))$ is compact by A Closed Euclidean Ball is Convex and Compact, hence totally bounded in $(\mathbb{R}^{n},d_{E})$ by A Compact Subset of a Metric Space is Totally Bounded; by that definition, read with the positive real number $s$, there is a finite set $F_{m,s}\subseteq\mathbb{R}^{n}$ with

In every other case put $F_{m,s}=\emptyset$. Let $D=\bigcup_{m\in\mathbb{N}}\bigcup_{s\in\mathbb{Q}}F_{m,s}$.

2. $D$ is countable. A finite subset $F$ of $\mathbb{R}^{n}$ is countable: either $F=\emptyset$, or by Finite Set there are $p\in\mathbb{N}$ and a bijection $j\mapsto a_{j}$ from the initial segment $[p]$ onto $F$, and then $F$ is the set of terms of the sequence whose $j$th term is $a_{j}$ when $j\le p$ and $a_{p}$ when $p<j$, these two cases being exhaustive by claim 3 of Properties of the Order on the Natural Numbers. The set $\mathbb{Q}$ is countable by The Integers and the Rational Numbers are Countable and nonempty, so by Countable Set it is the set of terms of a sequence $(s_{l})_{l\in\mathbb{N}}$. Hence for each $m\in\mathbb{N}$ the set $\bigcup_{s\in\mathbb{Q}}F_{m,s}=\bigcup_{l\in\mathbb{N}}F_{m,s_{l}}$ is countable by A Countable Union of Countable Sets is Countable, and $D$, the union over $m\in\mathbb{N}$ of these sets, is countable by the same lemma.

3. $D$ is dense. Let $x\in\mathbb{R}^{n}$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 1 of The Archimedean Property of the Real Numbers there is $m\in\mathbb{N}$ with $\lVert x\rVert<\iota(m)$; since $0\le\lVert x\rVert$ by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) gives $0<\iota(m)$. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and the symmetry of $d_{E}$ (Metric Space) we have $d_{E}(0,x)=\lVert x\rVert<\iota(m)$, so $x\in\bar{B}(0,\iota(m))$ by Closed Ball in a Metric Space. By The Rational Numbers are Dense in the Real Numbers there is $s\in\mathbb{Q}$ with $0<s<\varepsilon$. The set $F_{m,s}$ was therefore produced by step 1, so there is $a\in F_{m,s}\subseteq D$ with $x\in B(a,s)$, that is $d_{E}(a,x)<s$; by symmetry of $d_{E}$ and mixed transitivity, $d_{E}(x,a)<\varepsilon$. As $\varepsilon$ was arbitrary, claim 3 of Characterization of the Closure in a Metric Space by Open Balls gives $x\in\operatorname{cl}_{\mathbb{R}^{n}}(D)$. Hence $\operatorname{cl}_{\mathbb{R}^{n}}(D)=\mathbb{R}^{n}$ and $D$ is dense in $\mathbb{R}^{n}$.

By steps 2 and 3 the set $D$ is countable and dense, so $(\mathbb{R}^{n},d_{E})$ is separable. This is claim 1 of the statement.