The results: :Compact Space is Countably Compact :Countably Compact Metric Space is Sequentially Compact show that it suffices to prove that $M$ is compact. {{AimForCont}} that $M$ is not compact. Let $\CC$ be an open cover for $A$ such that $\CC$ does not have a finite subcover for $A$. By the definition of a totally bounded metric space, we can use the axiom of countable choice to obtain a sequence $\sequence {F_n}_{n \mathop \in \N}$ such that: :For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$. For the sake of notational brevity, let $\map {B'_n} a$ denote the open $2^{-n}$-ball of $a$ in $M$. For all $n \in \N$, define: :$G_n = \leftset {y \in F_n: \map {B'_n} y}$ is not covered by any finite subset of $\rightset \CC$ Since $F_n$ is finite by definition, it follows by the definition of a net that $G_n$ is non-empty. For all $y \in G_n$, define: :$\map {T_n} y = \leftset {z \in G_{n + 1}: \map {B'_n} y \cap \map {B'_{n + 1} } z}$ is not covered by any finite subset of $\rightset \CC$ By the definition of a net, it follows from the distributivity of intersection over union that: :$\ds \map {B'_n} y = \bigcup_{z \mathop \in F_{n + 1} } \paren {\map {B'_n} y \cap \map {B'_{n + 1} } z}$ Hence, by the definition of $G_n$, it follows that $\map {T_n} y$ is non-empty. From Countable Union of Countable Sets is Countable, it follows that the disjoint union $\ds \bigsqcup_{n \mathop \in \N} G_n$ is countable. Using the axiom of countable choice, we can obtain a sequence $\sequence {\phi_n: G_n \to G_{n + 1} }_{n \mathop \in \N}$ of mappings such that: :$\forall n \in \N: \forall y \in G_n: \map {\phi_n} y \in \map {T_n} y$ Let $x_0 \in G_0$. For all $n \in \N$, define: :$x_{n + 1} = \map {\paren {\phi_n \circ \cdots \circ \phi_1 \circ \phi_0} } {x_0}$ where $\circ$ denotes composition of mappings. For all $n \in \N$, define: :$A_n = \map {B'_n} {x_n} \cap \map {B'_{n + 1} } {x_{n + 1} }$ Note that $A_n$ is non-empty; otherwise, by Union of Empty Set, $\O$ would be a cover for $A_n$. Let $y \in A_n$. Then: {{begin-eqn}} {{eqn | l = \map d {x_n, x_{n + 1} }       | o = \le       | r = \map d {x_n, y} + \map d {x_{n + 1}, y}       | c =  }} {{eqn | o = <       | r = \frac 1 {2^n} + \frac 1 {2^{n + 1} }       | c =  }} {{eqn | r = \frac 3 {2^{n + 1} }       | c =  }} {{end-eqn}} Let $i < j$, where $i, j \in \N$. Then, by Sum of Infinite Geometric Progression: {{begin-eqn}} {{eqn | l = \map d {x_i, x_j}       | o = \le       | r = \sum_{k \mathop = i}^{j - 1} \map d {x_k, x_{k + 1} }       | c =  }} {{eqn | o = <       | r = \sum_{k \mathop = i}^\infty \frac 3 {2^{k + 1} }       | c =  }} {{eqn | r = \frac 3 {2^i}       | c =  }} {{end-eqn}} Hence, by Sequence of Powers of Number less than One, the sequence $\sequence {x_k}$ is Cauchy. By the assumption that $M$ is complete, the sequence $\sequence {x_k}$ converges to some limit $x \in A$. Choose $U \in \CC$ such that $x \in U$ (which can be done because $\CC$ covers $A$). By the definition of an open set, we can choose a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$. By Sequence of Powers of Number less than One, we can choose a natural number $n$ such that: :$\dfrac 1 {2^n} < \dfrac \epsilon 2$ By the definition of a limit, we can choose a natural number $m > n$ such that: :$\map d {x_m, x} < \dfrac 1 {2^n}$ For all $y \in A_m$, we have: {{begin-eqn}} {{eqn | l = \map d {y, x}       | o = \le       | r = \map d {y, x_m} + \map d {x_m, x}       | c =  }} {{eqn | o = <       | r = \dfrac 1 {2^m} + \dfrac 1 {2^n}       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{end-eqn}} That is: :$A_m \subseteq \map {B_\epsilon} x \subseteq U$ Since $\subseteq$ is a transitive relation, it follows that $A_m \subseteq U$. That is, $A_m$ is covered by the singleton $\set U \subseteq \CC$. But this contradicts the definitions of $\map {T_m} {x_m}$ and $\phi_m$. Hence our initial assumption that $M$ is not compact was false. Hence the result. {{qed}}