Here $\lambda^*$ is the Lebesgue outer measure.
Possible proof:
Assume first that $\lambda^*(E)$ is finite.
By the definition of Lebesgue outer measure, there exists sequence of open intervals $\{I_n\}$ s.t $E\subset \cup_n I_n$ and $\lambda^*(\cup_{n} I_n) \leq \sum l(I_n)< \frac{1}{\alpha}\lambda^*(E)$ where $l$ is the length function.
Suppose for contradiction that all $I_n$ do not meet the required condition in the question.
Hence, for all $n$: $\lambda^*(E\cap I_n)< \alpha \lambda^*( I_n)$
This implies that $ \lambda^*(E)=\lambda^*(E\cap \cup_n I_n)\leq \sum \lambda^*(E\cap I_n)\leq \sum \alpha \lambda^*( I_n)= \alpha\sum l( I_n) $. This is a contradiction according to the definition of $I_n$.
If $\lambda^*(E)$ is infinite that we can take $I$ to be R
Is this proof valid?