Proof of A Totally Bounded Metric Space is Separable

Let $\mathbb{N}$ denote the natural numbers and let $\iota:\mathbb{N}\to\mathbb{R}$ be the canonical map of the natural numbers into the real numbers. Write $B_d$ for the open ball in $(X,d)$.

Claim 2. Suppose first that $X$ is sequentially compact in $(X,d)$. Applying Compactness and Sequential Compactness Agree for Subsets of a Metric Space with the subset $K=X$ shows that $X$ is compact in $(X,\mathcal{T}_d)$. Suppose now that $X$ is compact in $(X,\mathcal{T}_d)$. Applying A Compact Subset of a Metric Space is Totally Bounded with the subset $K=X$ shows that $X$ is totally bounded in $(X,d)$. The final assertion of claim 2 is then claim 1 applied to $X$.

Claim 1. Assume $X$ is totally bounded in $(X,d)$.

First, for every $n\in\mathbb{N}$ we have $0<\iota(n)$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; in particular $\iota(n)$ is not $0$, so its multiplicative inverse $\iota(n)^{-1}$ exists in the ordered field of real numbers, and $0<\iota(n)^{-1}$ by claim 7 of Elementary Order Arithmetic in an Ordered Field.

Let $S$ be the set whose elements are the subsets of $X$. For $n\in\mathbb{N}$ let $\mathcal{A}_n$ be the set of those $F\in S$ that are finite and satisfy

Each $\mathcal{A}_n$ is a subset of $S$, and each is nonempty: $\iota(n)^{-1}$ is a real number with $0<\iota(n)^{-1}$, so total boundedness of $X$ in $(X,d)$ applied with $\varepsilon=\iota(n)^{-1}$ provides a finite subset of $X$ with exactly this property. By Axiom of Countable Choice, applied to the set $S$ and the family $(\mathcal{A}_n)_{n\in\mathbb{N}}$ of nonempty subsets of $S$, there is a sequence $(F_n)_{n\in\mathbb{N}}$ in $S$ with $F_n\in\mathcal{A}_n$ for every $n\in\mathbb{N}$.

Put

Each $F_n$ is a finite subset of $X$, hence countable by claim 2 of Basic Properties of Countable Sets, and $(F_n)_{n\in\mathbb{N}}$ is a family of subsets of $X$. Therefore $D$ is a countable subset of $X$ by A Countable Union of Countable Sets is Countable.

It remains to show that $D$ is dense. Let $x\in X$ and let $\varepsilon$ be a real number with $0<\varepsilon$. By claim 3 of The Archimedean Property of the Real Numbers there is $n\in\mathbb{N}$ such that $\iota(n)^{-1}$ exists and $0<\iota(n)^{-1}<\varepsilon$. Since $F_n\in\mathcal{A}_n$ we have

so, as $x\in X$, there is $a\in F_n$ with $x\in B_d(a,\iota(n)^{-1})$, that is, $d(a,x)<\iota(n)^{-1}$. By the symmetry axiom of a metric, $d(x,a)=d(a,x)<\iota(n)^{-1}$. From $\iota(n)^{-1}<\varepsilon$ we get $\iota(n)^{-1}\le\varepsilon$, so claim 2 of Elementary Order Arithmetic in an Ordered Field, applied to $d(x,a)<\iota(n)^{-1}$ and $\iota(n)^{-1}\le\varepsilon$, gives $d(x,a)<\varepsilon$. Finally $a\in F_n\subseteq D$.

Thus for every real $\varepsilon>0$ there is an element of $D$ at distance less than $\varepsilon$ from $x$, which is condition 3 of Characterization of the Closure in a Metric Space by Open Balls for the subset $D$ at the point $x$; by that theorem $x$ belongs to the closure of $D$ in $X$. Since $x\in X$ was arbitrary, $X$ is contained in that closure, and the closure is a subset of $X$ by definition, so the closure of $D$ in $X$ equals $X$. Hence $D$ is dense in $X$ for $\mathcal{T}_d$, and since $D$ is countable this exhibits $(X,d)$ as separable.