A key definition here is "inner measure". The Lebesgue inner measure of a set $A$ is the supremum of the measures of the Lebesgue measurable subsets of $A$ (equivalently, of the closed subsets of $A$). You are asking whether every nonmeasurable set has positive inner measure, and as user8268 has shown, the answer is no.
A set of real numbers $A$ of finite outer measure is measurable if and only if its Lebesgue inner measure $m_*(A)$ is equal to its Lebesgue outer measure $m^*(A)$. If $m^*(A)=m_*(A)$, then there is a $G_\delta$ set $G$ and an $F_\sigma$ set $F$ such that $F\subseteq A\subseteq G$ and $m(F)=m_*(A)=m^*(A)=m(G)$. Then $A\setminus F\subseteq G\setminus F$ is a null set, and therefore $A=F\cup(A\setminus F)$ is measurable.
Let $A$ be a nonmeasurable set with finite inner measure. Let $F\subset A$ be an $F_\sigma$ such that $m(F)=m_*(A)$. Then $A\setminus F$ is a nonmeasurable set with inner measure $0$. Thus, every nonmeasurable set contains nonmeasurable subsets with no measurable subsets of positive measure.