= 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 $d_m\to0$ to choose increasing indices $k_j$ with
$$
\delta_j:=d_{k_j}>0,\qquad \delta_{j+1}\leq\delta_j/8.
$$
Set $y_j=x_{k_j}$. The preceding bounds give $|\delta_j-\delta_\ell|\leq d(y_j,y_\ell)\leq\delta_j+\delta_\ell$. For $m\geq2$,
$$
\frac{d(y_{m+1},y_m)}{d(y_m,y_{m-1})}\leq\frac{(9/8)\delta_m}{(7/8)\delta_{m-1}}\leq\frac9{56}<\frac13.
$$
For $m<n$, use $\delta_n\leq\delta_m/8$ and $\delta_{n+1}\leq\delta_m/64$ to obtain
$$
\frac{d(y_{m+1},y_{n+1})}{d(y_m,y_n)}\leq\frac{\delta_m/8+\delta_m/64}{(7/8)\delta_m}=\frac9{56}<\frac12.
$$
The same holds when $n<m$ by symmetry; when $m=n$, both distances are zero. This proves \b[both required inequalities]. The adjacent-distance inequality is understood for $m\geq2$, since this <subsequence> starts at index one and has no $y_0$.
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