You are asking if the Lebesgue measure has any atoms $E$, i.e., a minimal measurable set of positive measure. The answer is negative: the Lebesgue measure is atomless. [My answer assumes we're working with $\mathbb R$, but the main idea carries over for all dimensions.]
Proof 1. Fix any measurable $E$ of positive measure, and a real $r$ such that $0 < r < m(E)$. For $n \in \mathbb Z$, define $J_n := [n r, (n+1) r)$ and $E_n := E \cap J_n$. Thus the $J_n$'s (resp. $E_n$'s) partition $\mathbb R$ (resp. $E$) into sets of measure at most $r$.
- For any $n$, $E_n$ is a measurable subset of $E$.
- From $E_n \subseteq J_n$, we have $m(E_n) \leqslant m(J_n) = r < m(E)$. [In particular, $E_n$ has finite measure even if $m(E) = \infty$.]
- From subadditivity, we have $m(E) \leqslant \sum_n \ m(E_n)$. Thus, assuming $m(E) > 0$, there exists some $n$ such that $m(E_n) > 0$. Such an $E_n \subseteq E$ satisfies $0 < m(E_n) < m(E)$.
Proof 2. (Slightly modified from Robert Israel's comment.) Let $B_r$ denote the (open or closed) ball of radius of $r$ about the origin. Then the function $h : [0, \infty) \to [0, \infty)$ defined by $h(r) := m(B_r \cap E)$ is both monotonically increasing and continuous.* Further, $h(0) = 0$, and $h(r) \to m(E)$ as $r \to \infty$. Therefore, by the intermediate value theorem, we can conclude that for every $0 \leqslant t \lt m(E)$, there exists some $0 \leqslant r \lt \infty$ such that $h(r) = m(E \cap B_r) = t$.
*Proof left as exercise.
Edit. I rewrote my answer so as not to appeal to contradiction. I hope this is simpler. Thanks to Nate, GEdgar and Michael for their comments regarding the terminology. Thanks to Robert for his comment.