= Solution
The <intersection> $G$ is a <closed set> because arbitrary <intersections> of <closed sets> are closed. Its nonemptiness needs a construction: bounded closed sets need not be <compact sets> in this setting.
Fix $n$, $x_n\in F_n$, and $\eta>0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\geq n$, using the <approximation property of an infimum> so that
$$
d(x_k,x_{k+1})<2^{-k}+\eta\,2^{n-k-1}.
$$
This is possible because $x_k\in F_k$ and $\operatorname{dist}(x_k,F_{k+1})\leq\rho(F_k,F_{k+1})\leq2^{-k}$. The two bounds form convergent <geometric series>, so the <triangle inequality> shows $(x_k)_{k\geq n}$ is a <Cauchy sequence>. By <completeness> it converges to some $x\in X$.
For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne\varnothing$. Moreover,
$$
d(x_n,x)\leq\sum_{k=n}^{\infty}\bigl(2^{-k}+\eta\,2^{n-k-1}\bigr)=2^{1-n}+\eta.
$$
For each starting point $x_n$, let $\eta\downarrow0$ in this bound on its <distance to a set> $G$. This does not assert that a nearest point exists. It proves $\operatorname{dist}(x_n,G)\leq2^{1-n}$. Since $G\subseteq F_n$, the other directed <Hausdorff distance> is zero, and
$$
\boxed{G\in\mathcal F(X),\qquad\rho(F_n,G)\leq2^{1-n}\longrightarrow0}.
$$
This is the <Summably nested Hausdorff intersection theorem>; the summable movement replaces any need for <compactness>.
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