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 $n$.