In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness

Nestedness propagates: for $k\le l$ one has $F_l\subseteq F_k$, by induction on $l$ from $F_{l+1}\subseteq F_l$ and transitivity of inclusion.

Assume [A1] and let $(F_k)$ be a Cantor chain. Every $F_k$ is nonempty, so [L4] supplies a sequence $(x_k)$ with $x_k\in F_k$ for every $k$.

A preliminary about closures, used in claim 2: let $A\subseteq X$ be nonempty and bounded, let $u,v\in \bar{A}$ and let $\eta >0$ be real; then $B(u,\eta )$ and $B(v,\eta )$ meet $A$, so there are $a,b\in A$ with $d(u,a)<\eta$ and $d(v,b)<\eta$, whence $d(u,v)\le d(u,a)+d(a,b)+d(b,v)<\mathrm{diam}(A)+2\eta$.

If $x,y\in {\bigcap }_k{F}_k$ then $d(x,y)\le \mathrm{diam}(F_k)$ for every $k$ by [L1]; the constant sequence with value $d(x,y)$ converges to $d(x,y)$ and $\mathrm{diam}(F_k)\to 0$, so $d(x,y)\le 0$, and $d(x,y)\ge 0$ forces $d(x,y)=0$ and $x=y$.

For claim 2 assume [A2] and let $(x_k)$ be a Cauchy sequence in $X$; put $A_k:=\{x_j:j\ge k\}$ and $F_k:=\overline{A_k}$.

Since $d(u,v)<\mathrm{diam}(A)+2\eta$ for every real $\eta >0$, we get $d(u,v)\le \mathrm{diam}(A)$: were $d(u,v)>\mathrm{diam}(A)$, the value $\eta :=(d(u,v)-\mathrm{diam}(A))\mathrm{/}3$ would be positive and would give $d(u,v)<\mathrm{diam}(A)\mathrm{/}3+2d(u,v)\mathrm{/}3<d(u,v)$.

Back to claim 1: for any $K\in \mathbb{N}$ and all $m,n\ge K$ we have $x_m\in F_m\subseteq F_K$ and $x_n\in F_n\subseteq F_K$, so $d(x_m,x_n)\le \mathrm{diam}(F_K)$.

$A_{k+1}\subseteq A_k$, and $\overline{A_k}$ is a closed superset of $A_{k+1}$, so $F_{k+1}\subseteq F_k$ by minimality of the closure.

Hence $\mathrm{diam}(A)$ is an upper bound of $\{d(u,v):u,v\in \bar{A}\}$; fixing $u\in \bar{A}$, which exists since $A\mathrm{\ne }\mathrm{\varnothing }$ and $A\subseteq \bar{A}$, gives $\bar{A}\subseteq B(u,\mathrm{diam}(A)+1)$, so $\bar{A}$ is nonempty and bounded and $\mathrm{diam}(\bar{A})\le \mathrm{diam}(A)$. And $\{d(a,b):a,b\in A\}\subseteq \{d(u,v):u,v\in \bar{A}\}$ gives $\mathrm{diam}(A)\le \mathrm{diam}(\bar{A})$, so the two diameters are equal.

Given a real $\epsilon >0$, the convergence $\mathrm{diam}(F_k)\to 0$ supplies $K$ with $\mathrm{diam}(F_K)<\epsilon$, so $d(x_m,x_n)<\epsilon$ for all $m,n\ge K$; hence $(x_k)$ is Cauchy, and by [A1] it converges to some $x\in X$.

Fix $K\in \mathbb{N}$. For every $k\ge K$ we have $x_k\in F_k\subseteq F_K$, and the tail $(x_{K+j}{)}_{j\in \mathbb{N}}$ converges to $x$ because $(x_k)$ does; since $F_K$ is closed it is sequentially closed, so $x\in F_K$. As $K$ was arbitrary, $x\in {\bigcap }_k{F}_k$.

Each $A_k$ is nonempty and is contained in the bounded range of $(x_k)$, hence bounded; so each $F_k$ is nonempty, closed and, by step 3.1, bounded with $\mathrm{diam}(F_k)=\mathrm{diam}(A_k)$.

Claim 1 is established: the intersection contains $x$ by step 4.1 and no second point by step 1.4.

Given a real $\epsilon >0$, Cauchyness supplies $K$ with $d(x_m,x_n)<\epsilon \mathrm{/}2$ for all $m,n\ge K$; then $\epsilon \mathrm{/}2$ is an upper bound of $\{d(a,b):a,b\in A_k\}$ for every $k\ge K$, so $0\le \mathrm{diam}(A_k)\le \epsilon \mathrm{/}2<\epsilon$ for $k\ge K$. Hence $\mathrm{diam}(F_k)\to 0$ and $(F_k)$ is a Cantor chain.

By [A2] there is $x\in {\bigcap }_k{F}_k$. Given a real $\epsilon >0$, take $K$ as in step 5.2 for $\epsilon$; since $x\in F_K=\overline{A_K}$, the ball $B(x,\epsilon \mathrm{/}2)$ meets $A_K$, so there is $j\ge K$ with $d(x,x_j)<\epsilon \mathrm{/}2$, and then for every $k\ge K$ we get $d(x,x_k)\le d(x,x_j)+d(x_j,x_k)<\epsilon \mathrm{/}2+\epsilon \mathrm{/}2=\epsilon$.

So $x_k\to x$ with $x\in X$, every Cauchy sequence in $X$ converges, and $(X,d)$ is complete; this is claim 2, and claim 1 is step 5.1.