= Solution
This requires an argument using <Lebesgue outer measure>, since $E$ need not be a <Lebesgue measurable set>. Some bounded portion $F=E\cap[-N,N]$ has $a=m^*(F)>0$: otherwise <countable subadditivity> would give $m^*(E)=0$. Also $a\leq2N<\infty$.
Suppose, for a contradiction, that every nonempty bounded open <interval> $I$ satisfies $m^*(E\cap I)\leq\lambda|I|$. Because $a/\lambda>a$, the definition of <Lebesgue outer measure> gives a countable cover $F\subseteq\bigcup_j I_j$ by nonempty bounded open intervals with
$$
\sum_j|I_j|<\frac a\lambda.
$$
Using <countable subadditivity> again,
$$
a=m^*(F)\leq\sum_jm^*(F\cap I_j)\leq\sum_jm^*(E\cap I_j)\leq\lambda\sum_j|I_j|<a,
$$
a contradiction. Hence at least one interval in such a cover satisfies
$$
\boxed{m^*(E\cap I)>\lambda|I|.}
$$
This <high-density interval for outer measure> applies even to nonmeasurable sets; the argument does not require the <Lebesgue density theorem>.
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