Assume I have a sequence $\{a\_m\}$ that is vanishing and strictly positive:
$$
0<a\_{m+1}\leq a\_m\leq\ldots\leq a\_1<\infty, \quad \lim\_{m\to \infty}a\_m = 0
$$
Is it true or false that this has a subsequence $\{a\_{m\_n}\}$ such that
$$
a\_{m\_n}\asymp n^{-1}
$$
which is to say, there exists a constant $C>0$ such that for all $n$:
$$
\frac{1}{Cn} \leq a\_{m\_n} \leq \frac{C}{n}
$$



Vanishing sequence and subsequence with particular decay
The answer is no. E.g., let $a\_m:=1/(m!)^2$ and $n\_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na\_{m\_n}|$ is bounded. In our case, we have 
$$a\_{k+1}<\frac1{n\_k}<a\_k$$
for all $k$. 
So, for all $m\le k$ we have $1<n\_k a\_k\le n\_k a\_m$ and hence $|\ln(n\_k a\_m)|\ge|\ln(n\_k a\_k)|=\ln(k+1)$. Similarly, for all $m\ge k+1$ we have 
$1>n\_k a\_{k+1}\ge n\_k a\_m$ and hence $|\ln(n\_k a\_m)|\ge|\ln(n\_k a\_{k+1})|=\ln(k+1)$. So, $|\ln(n\_k a\_m)|\ge\ln(k+1)$ 
for all $k,m$. 


So, for any $(m\_n)$
$$|\ln n\_ka\_{m\_{n\_k}}|\ge\inf\_m|\ln n\_k a\_m|\ge\ln(k+1)\to\infty$$
as $k\to\infty$, so that $|\ln na\_{m\_n}|$ is unbounded, for any choice of $(m\_n)$. 


