 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 $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ 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,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.