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 $u, v \in \overline{A}$ 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 $d(u,v) \le d(u,a) + d(a,b) + d(b,v) < \operatorname{diam}(A) + 2\eta$.

If $x, y \in \bigcap_k F_k$ then $d(x,y) \le \operatorname{diam}(F_k)$ for every $k$ by [L1]; the constant sequence with value $d(x,y)$ converges to $d(x,y)$ and $\operatorname{diam}(F_k) \to 0$, 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 $d(u,v) < \operatorname{diam}(A) + 2\eta$ for every real $\eta > 0$, we get $d(u,v) \le \operatorname{diam}(A)$: were $d(u,v) > \operatorname{diam}(A)$, the value $\eta := (d(u,v) - \operatorname{diam}(A))/3$ would be positive and would give $d(u,v) < \operatorname{diam}(A)/3 + 2d(u,v)/3 < d(u,v)$.

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 $d(x_m,x_n) \le \operatorname{diam}(F_K)$.

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

Given a real $\varepsilon > 0$, the convergence $\operatorname{diam}(F_k) \to 0$ supplies $K$ with $\operatorname{diam}(F_K) < \varepsilon$, so $d(x_m,x_n) < \varepsilon$ 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 $\operatorname{diam}(F_k) = \operatorname{diam}(A_k)$.

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

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

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

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.