Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences

The quadrilateral estimate of [L1] gives $|d(x_n,y_n) - d(x_m,y_m)| \le d(x_n,x_m) + d(y_n,y_m)$ for all $n,m$.

Given a real $\varepsilon > 0$, [A1] supplies $K_x$ and $K_y$ with $d(x_n,x_m) < \varepsilon/2$ and $d(y_n,y_m) < \varepsilon/2$ for indices beyond them; with $K := \max\{K_x,K_y\}$ the sequence $\big(d(x_n,y_n)\big)_n$ satisfies $|d(x_n,y_n) - d(x_m,y_m)| < \varepsilon$ for $n,m \ge K$, so it is a Cauchy sequence of reals.

By [L2] that sequence converges and its limit is unique, so $\rho(x,y) := \lim_n d(x_n,y_n)$ is a single well-determined real: claim 1.

$\rho$ is nonnegative and symmetric, and satisfies $\rho(x,z) \le \rho(x,y) + \rho(y,z)$: the terms $d(x_n,y_n)$ are nonnegative, $d(x_n,y_n) = d(y_n,x_n)$, and $d(x_n,z_n) \le d(x_n,y_n) + d(y_n,z_n)$ for every $n$, and all three pass to the limit.

Claim 2: $\sim$ is reflexive since $d(x_n,x_n) = 0$ for every $n$; symmetric since $\rho$ is; and transitive, since $\rho(x,y) = \rho(y,z) = 0$ gives $0 \le \rho(x,z) \le 0 + 0 = 0$. So $\sim$ is an equivalence relation and $\widehat{X} = \mathcal{C}/\!\sim$ is defined.

Claim 3, well-definedness: if $x \sim x'$ and $y \sim y'$ then, by [L1] applied termwise, $|d(x_n,y_n) - d(x'_n,y'_n)| \le d(x_n,x'_n) + d(y_n,y'_n)$; passing to the limit gives $|\rho(x,y) - \rho(x',y')| \le \rho(x,x') + \rho(y,y') = 0$, so $\rho(x,y) = \rho(x',y')$. Hence $\widehat{d}([x],[y]) := \rho(x,y)$ is a well-defined function on $\widehat{X} \times \widehat{X}$.

$\widehat{d}$ is a metric: symmetry and the triangle inequality are step 4.1 read on classes, and $\widehat{d}([x],[y]) = 0$ says $\rho(x,y) = 0$, which says $x \sim y$, which says $[x] = [y]$. This completes claim 3.

Claim 4: for $p \in X$ the constant sequence at $p$ is Cauchy, so $\iota(p) := [(p)_n]$ is defined, and $\widehat{d}(\iota(p),\iota(q)) = \lim_n d(p,q) = d(p,q)$, a constant sequence; so $\iota$ is an isometric embedding.

Density: let $[x] \in \widehat{X}$ and let $\varepsilon > 0$ be real. By [A1] there is $N$ with $d(x_n,x_m) < \varepsilon/2$ for all $n,m \ge N$; in particular $d(x_n,x_N) \le \varepsilon/2$ for all $n \ge N$, so $\widehat{d}([x], \iota(x_N)) = \lim_n d(x_n,x_N) \le \varepsilon/2 < \varepsilon$. Hence every ball around $[x]$ meets $\iota[X]$, that is $\iota[X]$ is dense, completing claim 4.

Claim 5: let $(\xi^{(k)})$ be a Cauchy sequence in $(\widehat{X},\widehat{d})$. For each $k$ the set $A_k := \{\, p \in X : \widehat{d}(\xi^{(k)}, \iota(p)) < 1/(k+1) \,\}$ is nonempty by step 9.1, so [L6] supplies $a_k \in A_k$ for every $k$, that is a sequence $(a_k)$ in $X$ with $\widehat{d}(\xi^{(k)}, \iota(a_k)) < 1/(k+1)$.

$(a_k)$ is Cauchy in $X$: since $\iota$ is isometric, $d(a_k,a_l) = \widehat{d}(\iota(a_k),\iota(a_l)) \le 1/(k+1) + \widehat{d}(\xi^{(k)},\xi^{(l)}) + 1/(l+1)$; given a real $\varepsilon > 0$, choose $K$ so large that $1/(K+1) < \varepsilon/3$ and $\widehat{d}(\xi^{(k)},\xi^{(l)}) < \varepsilon/3$ for all $k,l \ge K$, and then $d(a_k,a_l) < \varepsilon$ for all $k,l \ge K$. So $a := (a_k) \in \mathcal{C}$ and $[a] \in \widehat{X}$.

$\xi^{(k)} \to [a]$: given a real $\varepsilon > 0$ take $K$ as in step 11.1 for $\varepsilon/3$, so that $d(a_k,a_n) < \varepsilon/3$ for all $k,n \ge K$, enlarged if necessary so that also $1/(K+1) < \varepsilon/3$. For $k \ge K$ we have $d(a_k,a_n) \le \varepsilon/3$ for all $n \ge K$, hence $\widehat{d}(\iota(a_k),[a]) = \lim_n d(a_k,a_n) \le \varepsilon/3$, and therefore $\widehat{d}(\xi^{(k)},[a]) \le \widehat{d}(\xi^{(k)},\iota(a_k)) + \widehat{d}(\iota(a_k),[a]) < \varepsilon/3 + \varepsilon/3 < \varepsilon$.

So every Cauchy sequence in $(\widehat{X},\widehat{d})$ converges in it, which is claim 5; with claims 1 to 4 this exhibits $\big((\widehat{X},\widehat{d}),\iota\big)$ as a completion of the arbitrary metric space $(X,d)$.