On $\mathbb{N}$ with $d(m,n)=1+1\mathrm{/}(m+n)$ for $m\mathrm{\ne }n$ the sets $\{n,n+1,\dots \}$ are nested, closed, bounded and complete with empty intersection

$d$ is well defined and symmetric, and $d(m,n)=0$ exactly when $m=n$, since for $m\mathrm{\ne }n$ the value $1+1\mathrm{/}(m+n)$ is at least $1$. In fact $1<d(m,n)\le 2$ for all $m\mathrm{\ne }n$, because $0<1\mathrm{/}(m+n)\le 1$.

$d$ satisfies the triangle inequality $d(m,p)\le d(m,n)+d(n,p)$. If $m=p$ the left side is $0$; if $n=m$ or $n=p$ one side of the right-hand sum is $0$ and the other equals the left side; and if $m$, $n$, $p$ are pairwise distinct then $d(m,n)+d(n,p)>1+1=2\ge d(m,p)$. So $d$ is a metric on $\mathbb{N}$.

Every subset of $\mathbb{N}$ is open, hence also closed: for $m\in \mathbb{N}$ the ball $B(m,1)$ is $\{m\}$, since $d(m,n)>1$ for $n\mathrm{\ne }m$, so every subset is a union of open balls; complements of subsets are subsets.

$(X,d)$ is complete: a Cauchy sequence $(x_k)$ tested at $\epsilon =1$ has an index $K$ with $d(x_k,x_l)<1$ for all $k,l\ge K$, which by step 1.1 forces $x_k=x_l$; so the sequence is constant from $K$ on and converges to $x_K\in \mathbb{N}$.

Each $F_n$ is bounded: $d(0,k)\le 2<3$ for every $k$, so $F_n\subseteq X=B(0,3)$.

Each $F_n$ is nonempty (it contains $n$), closed by step 2.2, and complete by [L6]; and $F_{n+1}\subseteq F_n$, so the family is nested.

$\mathrm{diam}(F_n)=1+1\mathrm{/}(2n+1)$. Indeed the distances realised inside $F_n$ are $0$ and the values $1+1\mathrm{/}(j+k)$ for distinct $j,k\ge n$; among such pairs $j+k$ is least when $\{j,k\}=\{n,n+1\}$, giving $j+k=2n+1$, and $1\mathrm{/}(j+k)$ is largest there. So $1+1\mathrm{/}(2n+1)$ is an upper bound of those distances and is itself one of them, hence it is the least upper bound.

In particular $\mathrm{diam}(F_n)>1$ for every $n$, so the sequence $(\mathrm{diam}(F_n))$ does not converge to $0$: at $\epsilon =1$ no index makes the diameters smaller than $1$.

The intersection is empty: for every $k\in \mathbb{N}$ we have $k\mathrm{\notin }{F}_{k+1}$, since $k<k+1$; so no natural lies in all the $F_n$.

So in the complete metric space $(X,d)$ the sets $F_n$ are nonempty, closed, bounded, complete and nested, and their intersection is empty; the only hypothesis of In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness that they fail is $\mathrm{diam}(F_n)\to 0$, so that hypothesis cannot be dropped.