This is a generalization of an older question.
Let $(X,d)$ be a metric space and let $(A_n)_{n\in\mathbb{N}}\subseteq X$ be a sequence of non-empty closed subsets such that for all $\varepsilon > 0$ there is $N\in\mathbb{N}$ such that for all $m,n\geq N$ we have $d_H(A_m,A_n)<\varepsilon$ (where $d_H(\cdot,\cdot)$ denotes the Hausdorff distance).
Can we pick $x_n\in A_n$ for all $n\in\mathbb{N}$ such that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence?
Yes. Choosing a subsequence $n_1<n_2<\dots$ such that $d_H(A_{n_i},A_{n_i+1})\leqslant 2^{-i}$. Then inductively choose points $x_{n_i}\in A_{n_i}$ so that $d(x_{n_i},x_{n_{i+1}})\leqslant 2^{-i}$. It is Cauchy sequence by triangle inequality. For any index $m\notin \{n_1,n_2,\dots\}$ choose any $i$ such that $n_i>m$ and a point $x_m\in A_m$ on the distance at most $d_H(A_m,A_{n_i})+1/m$ from $x_{n_i}$. The whole sequence $(x_1,x_2,\dots)$ is still Cauchy.
