 Solution

First handle possible zero values. If $d_m=d_n=0$, the previous upper bound forces $x_m=x_n$. Thus, if infinitely many $d_m$ vanish, choosing those indices gives a constant subsequence, which satisfies both inequalities.

Otherwise, discard the finitely many zero values and use $d_m\to 0$ to choose increasing indices $k_j$ withSet $y_j=x_{k_j}$. The preceding bounds give $|{\delta }_j-{\delta }_{\ell }|\le d(y_j,y_{\ell })\le {\delta }_j+{\delta }_{\ell }$. For $m\ge 2$,For $m<n$, use ${\delta }_n\le {\delta }_m/8$ and ${\delta }_{n+1}\le {\delta }_m/64$ to obtainThe same holds when $n<m$ by symmetry; when $m=n$, both distances are zero. This proves both required inequalities. The adjacent-distance inequality is understood for $m\ge 2$, since this subsequence starts at index one and has no $y_0$.