For set of positive measure $E$, $\alpha \in (0, 1)$, there is interval $I$ such that $m(E \cap I) > \alpha \, m(I)$

@AymanHourieh's answer is simple and direct. I will add this slightly longer (yet perhaps more basic) solution that does not rely on Lebesgue's density theorem.

We start with a basic claim on measurable sets (which is sometimes given as the definition for a measurable set):

Claim. Let $E \in M$ of finite measure, then for every $\epsilon>0$ there exists a finite union of disjoint intervals: $$A_\epsilon := \biguplus_{n=1}^{N}I_n$$ s.t. $m(E \triangle A_\epsilon) < \epsilon$.

Now, let $\epsilon:=(1-\alpha)m(E)$. And let $A_\epsilon$ from the claim. Then $$ m(E) = m(E\cap A_\epsilon) + m(E\setminus A_\epsilon) \\ \leq m(E\cap A_\epsilon) + m(E \triangle A_\epsilon) \\ < m(E\cap A_\epsilon) + \epsilon \\ = m(E\cap A_\epsilon) + (1-\alpha)m(E) $$ Then $$ m(E) < \frac{m(E\cap A_\epsilon)}{\alpha} \tag{1} $$ Similarly, $$ m(A_\epsilon) = m(E\cap A_\epsilon) + m(A_\epsilon \setminus E) \\ \leq m(E\cap A_\epsilon) + (1-\alpha)m(E) \\ \overset{\text{from (1)}}{<} m(E\cap A_\epsilon) + \frac{1-\alpha}{\alpha}m(E\cap A_\epsilon) \\ = \frac{m(E\cap A_\epsilon)}{\alpha} $$ We now note that $$ m(A_\epsilon ) = \sum_{n=1}^{N}m(I_n) \\ m(E \cap A_\epsilon ) = \sum_{n=1}^{N}m(E \cap I_n). $$ Therefore, $$ \sum_{n=1}^{N}m(I_n) < \frac{1}{\alpha}\sum_{n=1}^{N}m(E \cap I_n), $$ So there must exist $1\leq n \leq N$ s.t. $$ m(I_n) < \frac{1}{\alpha}m(E \cap I_n), $$ which completes the proof.

Here is an (simpler?) solution:

Suppose not. Then for each interval $I$ we have $m(E \cap I) \leq \alpha m(I)$. Let $\varepsilon > 0$. By the definition of the Lebesgue measure, there is some open set $G \supseteq E$ such that $m(G) < m(E) + \varepsilon$. Since every open subset of $\mathbb{R}$ is a (at most) countable union of open, disjoint intervals, we have $G = \bigcup_{k=1}^\infty (a_k,b_k)$. But then $$ m(E) = m(E \cap G) =\sum_{k=1}^\infty m(E \cap (a_k,b_k)) \leq \alpha \sum_{k=1}^\infty m((a_k,b_k)) = \alpha m(G) < \alpha (m(E)+\varepsilon) $$ Since $\varepsilon > 0$ was arbitrary, we have $m(E) \leq \alpha m(E)$. But then $m(E) = 0$ since $m(E) \geq 0$ and $0 < \alpha < 1$.

The Lebesgue density theorem states that for almost every point $x \in E$ with $m(E) > 0$, $$ \lim_{\epsilon\to 0} \frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} = 1. $$

Fix such $x$ and pick $\epsilon$ such that $$ \frac{m(E \cap B_\epsilon(x))}{m(B_\epsilon(x))} > \alpha. $$

$I = B_\epsilon(x)$ is the desired interval.