Does there exist a measurable subset $T$ of $[0, \infty)$ with finite measure and some $\epsilon > 0$ such that for every $r$ with $0 < r < \epsilon$, $nr$ is in $T$ for infinitely many positive integers $n$?
Note: The integers $n$ such that $nr$ lie in $T$ can depend on $r$.
No. Denote $T_k=T\cap [k,k+1)$. Then $\sum |T_k|<\infty$ (where $|X|$ stands for the measure of $X\subset \mathbb{R}$). Choose a segment $[a,b]\subset (0,\epsilon)$. Note that if $r\in [a,b]$ and $nr\in T_k$, then $na\leqslant nr< k+1$ and $nb\geqslant nr\geqslant k$, thus $n\in [k/b,(k+1)/a]$. The union of $n^{-1}T_k$ over all positive integers $n\in [k/b,(k+1)/a]$ has measure at most $$|T_k|\cdot \sum_{n\in [k/b,(k+1)/a]} n^{-1}\leqslant C|T_k|,$$ where $C$ depends only on $a$ and $b$. Now choose $k_0$ so that $C\sum_{k>k_0} |T_k|<b-a$. By the pigeonhole principle there exists a point $r\in [a,b]$ not covered by $n^{-1} T_k$ with $k>k_0$ and positive integer $n$.
