For some bounded interval $I$, start with some compact $K \subset I$ of positive measure that contains no interval (i.e, ${m(K \cap J) < m(J)}$ for every interval $J$). The complement $I \setminus K$ is open, and thus can be expressed as an at most countable union of disjoint open intervals $V_n$. For each $V_n$, let $K_n \subset V_n$ be a compact subset containing no interval with $m(K_n) > m(V_n)/2$. Define $E_1 := \bigcup_{n=1}^\infty K_n \subset I \setminus K$. Define $E_2 \subset (I \setminus K) \setminus E_1$ similarly and so on.
Let $E := \bigcup_{n=1}^\infty E_n$. For any subinterval $I' \subset I$, we can show that $m(E \cap I') > 0$. Yet I'm not sure about the other half of the inequality $m(E \cap I') < m(I')$. Or could it be that the $E$ constructed in this way does not satisfy the given property?