Let $\epsilon \in \R_{>0}$ be a strictly positive real number. By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$. We use a Proof by Contradiction. That is, suppose that there exists no finite $\epsilon$-net for $M$. The aim is to construct an infinite sequence $\sequence {x_n}_{n \ge 1}$ in $A$ that has no convergent subsequence. For all natural numbers $n \ge 1$, define the set: :$\mathcal S_n = \set{F \subseteq A: \size F = n: \forall x, y \in F: x \ne y \implies \map d {x, y} \ge \epsilon}$ where $\size F$ denotes the cardinality of $F$. We use the Principle of Mathematical Induction to prove that $\mathcal S_n$ is non-empty. It is vacuously true that any singleton $\set x \subseteq A$ is an element of $\mathcal S_1$. Since $A$ is non-empty by the definition of a metric space, it follows from Existence of Singleton Set that $\mathcal S_1$ is non-empty. Let $F \in \mathcal S_n$. By definition, $F$ is finite. So $F$ is not an $\epsilon$-net for $M$, by hypothesis. Hence, there exists an $x \in A$ such that: :$\forall y \in F: \map d {x, y} \ge \epsilon$ Note that, by axiom $\left({M1}\right)$ for a metric: :$x \notin F$ Consider the set: :$F' := F \cup \set x$. Then: :$\left\vert{F'}\right\vert = n + 1$ and by axiom $\paren{M3}$ for a metric: :$F' \in \mathcal S_{n+1}$ Thus, we have proven that $\mathcal S_n$ is non-empty for all natural numbers $n \ge 1$. Therefore, using the axiom of countable choice, we can obtain an infinite sequence $\sequence {F_n}_{n \ge 1}$ such that: : $\forall n \in \N_{\ge 1}: F_n \in \mathcal S_n$ From Countable Union of Countable Sets is Countable, there exists an injection: :$\displaystyle \phi: \bigcup_{n \mathop \ge 1} F_n \to \N$ We now construct an infinite sequence $\sequence {x_n}_{n \ge 1}$ in $A$. {{explain|This seems a colloquial sense of PoRD, which needs to be formalised in some way}} To do this, we use the Principle of Recursive Definition to define the sequence $\sequence {\tuple {x_1, x_2, \ldots, x_n} }_{n \ge 1}$ of ordered $n$-tuples. Let $x_1 \in F_1$. Suppose that $x_1, x_2, \ldots, x_n$ have been defined, and let: :$T_n = \set {x_1, x_2, \ldots, x_n}$ Define: :$D_n = \set {x \in F_{n+1}: \forall y \in T_n: \map d {x, y} \ge \dfrac \epsilon 2}$ Using a Proof by Contradiction, we show that $D_n$ is non-empty. For all $x \in F_{n+1}$, define: :$\map {C_n} x = \set {y \in T_n: \map d {x, y} < \dfrac \epsilon 2}$ Let $x, x' \in F_{n+1}$ be distinct. Let $y \in \map {C_n} x$. Then it follows from: :the definition of $F_{n+1}$ :axioms $\paren {M2}$ and $\paren {M3}$ for a metric that: :$\map d {x', y} \ge \map d {x, x'} - \map d {x, y} > \dfrac \epsilon 2$ Hence, $y \notin \map {C_n} {x'}$. That is, the indexed family of sets: :$\sequence {\map {C_n} x}_{x \in F_{n + 1}}$ is pairwise disjoint. Suppose that $D_n$ is empty. That is: :$\forall x \in F_{n+1}: \map {C_n} x$ is non-empty Then, from Cardinality is Additive Function, Finite Union of Sets in Additive Function, and Cardinality of Subset of Finite Set, we have: :$\displaystyle \size {F_{n+1}} \le \sum_{x \mathop \in F_{n+1}} \size {\map {C_n} x} \le \size {T_n} < \size {F_{n+1}}$ which is a contradiction. {{explain|"Therefore ..." and explain what assumption was made so as to give rise to this contradiction.}} From the well-ordering principle, we have that $\struct {\N, \le}$ is a well-ordered set. Let $\le_{\phi}$ be the ordering induced by $\phi$. Then $\le_{\phi}$ is a well-ordering. We define $x_{n+1}$ as the (unique) $\le_{\phi}$-smallest element of $D_n$. By construction: :$\forall m, n \in \N_{>0}: m \le n \implies \map d {x_{n+1}, x_m} \ge \dfrac \epsilon 2$ Hence, by induction, it follows from axiom $\paren {M3}$ for a metric that: :$\forall m, n \in \N_{>0}: m \ne n \implies \map d {x_m, x_n} \ge \dfrac \epsilon 2$ Therefore, the sequence $\sequence {x_n}$ has no Cauchy subsequence. From Convergent Sequence in Metric Space is Cauchy Sequence, $\sequence {x_n}$ has no convergent subsequence either. Thus, by definition, $M$ is not sequentially compact. But this contradicts the original assumption that $M$ is sequentially compact. Thus the assumption that there exists no finite $\epsilon$-net for $M$ was false. Therefore, by definition, $M$ is totally bounded. Hence the result. {{qed}}