= Solution
Let $(F_n)$ be a <Cauchy sequence> for the <Hausdorff distance>, and put
$$
\delta_n=\sup_{p,q\geq n}\rho(F_p,F_q),\qquad A_n=\overline{\bigcup_{m\geq n}F_m}.
$$
The <bounded set> hypothesis makes $\delta_n$ finite, and the <Cauchy sequence> property gives $\delta_n\to0$. Each $A_n$ is nonempty and a <closed set>, with $A_{n+1}\subseteq A_n$.
We need estimates that survive taking the <closure>. For every nonempty $F$, the <distance to a set> is <Lipschitz continuous> with constant one: the <triangle inequality> gives $\operatorname{dist}(x,F)\leq d(x,y)+\operatorname{dist}(y,F)$, and interchanging $x,y$ proves
$$
|\operatorname{dist}(x,F)-\operatorname{dist}(y,F)|\leq d(x,y).
$$
For every $m\geq n$, points of $F_m$ have distance at most $\delta_n$ from $F_n$. The same holds throughout their union and its <closure>, by this <continuity>. Since $F_n\subseteq A_n$,
$$
\rho(F_n,A_n)\leq\delta_n.
$$
More generally, if $m\geq n$, every point in $\bigcup_{k\geq n}F_k$ is within $\delta_n$ of $F_m\subseteq A_m$. Passing to its <closure> again, and using $A_m\subseteq A_n$, gives
$$
\rho(A_n,A_m)\leq\delta_n.
$$
Choose strictly increasing indices $n_j$ with $\delta_{n_j}\leq2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in <Hausdorff distance> to the nonempty <closed set> $G=\bigcap_j A_{n_j}=\bigcap_n A_n$, with $\rho(A_{n_j},G)\leq2^{1-j}$.
For $n\geq n_j$, the <triangle inequality> gives the explicit full-sequence bound
$$
\rho(F_n,G)\leq\rho(F_n,F_{n_j})+\rho(F_{n_j},A_{n_j})+\rho(A_{n_j},G)
\leq2\delta_{n_j}+2^{1-j}\leq2^{2-j}.
$$
It tends to zero as $j\to\infty$. Therefore
$$
\boxed{X\text{ complete}\ \Longrightarrow\ (\mathcal F(X),\rho)\text{ complete}}.
$$
This proves <Completeness of the Hausdorff hyperspace> using the proposed tail closures, rather than assuming their diameters shrink.
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