The subset that $m(E \cap I) \geq \alpha m(I)$ has measure 1. I'm stucked in this question:
If $E$ is a Lebesgue measurable set contained in the interval $[0,1]$ and there is a $\alpha>0$ such that $m(E \cap I) \geq \alpha m(I)$ for all open intervals $I \in [0, 1]$ then $m(E) = 1$.
How should I proceed to prove this?
 A: We prove it by contradiction. Assume the following holds: $m(E)=1-\epsilon$  for some $\epsilon>0$. Clearly $m(E^c)=\epsilon$. Now you would actually like to choose some open interval $I$ such that $I$ is around $E^c$. If you could do that then $m(E\cap I) $ would be small and $\alpha m(I)$ would be large and that gives you the contradiction. But how you choose such$I$? It can be the case that $E^c$ is scattered around the whole interval $[0,1]$ so finding an open interval $I$ "near" $E^c$ cannot be achieved. Luckily we have a supertool to fix that, namely Littlewood's First Principle.

Littlewood's First Principle (LFP): Let $A\subset\mathbb{R}$ be Lebesgue measurable with $m(A)<\infty$ and let $\delta>0$. Then there exists a finite collection of disjoint open intervals $\{B_1,...,B_k\}$ such that $m(A\triangle \bigcup_{i=1}^kB_i)<\delta$.

The triangle means symmetric difference, i.e. $A\triangle B=[A\cup B]\setminus [A\cap B]$. The theorem as stated above works also if we would take $[0,1]$ instead of $\mathbb{R}$ and the open intervals that we will get will be in $[0,1]$. 
We use LFP for $E^c$. Let $\delta=\alpha\epsilon/2>0$, then we get  $\{B_1,...,B_k\}$ with the mentioned property. Define $O:=\bigcup_{i=1}^NB_i$. So we have $m(E^c \triangle O)<\delta$.
Note that we have $\alpha m(B_j)\leq m(E\cap B_j )$ for all $j\in \{1,...,k\}$. Since the sets $B_j$ are mutually disjoint we get:
\begin{align}
\alpha m(O) \leq m(E\cap O)
\end{align}
Now do some little algebra:
\begin{align}
m(O) = m( E\cap O) + m(E^c \cap O)
\end{align}
Furthermore:
\begin{align}
m(E^c\triangle O) = m(E^c \cup O) - m(E^c\cap O)
\end{align}
So:
\begin{align}
m(O) =  m( E\cap O)+m(E^c \cup O)-m(E^c\triangle O)
\end{align}
So now we get:
\begin{align}
\alpha ( m(E\cap O) +m(E^c \cup O)-m(E^c\triangle O)) &\leq m(E\cap O)\\
\alpha ( m(E^c \cup O)-m(E^c\triangle O)) &\leq (1-\alpha) m(E\cap O)
\end{align}
Note also that $m(E\cap O)\leq m(E^c\triangle O)$ so:
\begin{align}
\alpha (m(E^c \cup O)-m(E^c\triangle O)) &\leq (1-\alpha) m(E^c\triangle O)\\
\alpha m(E^c \cup O) &\leq m(E^c\triangle O)
\end{align}
We finally get:
\begin{align}
m(E^c \cup O) < \frac \epsilon 2
\end{align}
Remember that $\epsilon=m(E^c) \leq m(E^c\cup O)<\epsilon/2$, a contradiction. So $m(E)=1$.
A: Outline: (1) Show that $m(E\cap U)\ge\alpha m(U)$ for every open subset of $[0,1]$; (2) Use the approximation theorem for general Lebesgue measurable sets to show that $m(E\cap B)\ge \alpha m(B)$ for every Lebesgue measurable subset of $[0,1]$; (3) Consider what happens in the special cae $B=E^c$.
A: Following your comment about the Lebesgue Density Theorem, if you want to use it, you can argue as follows: 
It is clear that $0<\alpha <1.$ Consider $F:=[0,1]\setminus E.$ Since $I\cap (E\cup F)=(I\cap E)\cup (I\cap F)=I,$ we have $m(I\cap F)=m(I)-m(I\cap E)\le (1-\alpha )m(I).$ Then, $\frac{m(I\cap F)}{m(I)}\le 1-\alpha.$ Since this is true for all intervals $I\in [0,1],$ we conclude from the Lebesgue Density Theorem that $m(F)=0$ which now implies that $m(E)=1.$
