The availability of open sets in non-zero Lebesgue measure sets (Exercise 31, Chapter 1 in Folland) I'm working on the following problem in Folland:

If $E \in \mathcal{L}$ and $m(E)>0$, the set $E - E = \{x-y:x,y\in E\}$ contains an interval centered at zero. (If $I$ is in the previous exercise with $\alpha > 3/4$, then $E-E$ contains $\big(- \frac{1}{2}m(I),\frac{1}{2}m(I)\big)$)

The previous exercise reads:

If $E \in \mathcal{L}$ and $m(E)>0$, for an $\alpha>0$ there is an open interval $I$ such that $m(E \cap I) > \alpha m(I)$,

I have a solution. But I'm questioning whether its correct, because I haven't used Folland's hint.
My potential solution follows: If I know I can find an open interval in any Lebesgue set of non zero measure, suppose I find one call it $(a,b)$. Then check 4 cases for the set $I-I \subset E-E$:
Case 1: $a>0, b<0$. In this case chose $(-\min(a,|b|),\min(a,|b|)) \subset I-I$
Case 2: $a>0, b>0$. $I-I = (b-a,a-b)$. Evidently the upper and lower bounds are positive and negative, respectively. So choose $\big(-\min(a-b,|b-a|),\min(a-b,|b-a|)\big)$
Case 3 and 4 follow similarly.


*

*Is what I'm saying correct? I can just construct these manually right?

*If this is so, what did Folland want me to see with the hint then?

 A: 
If I know I can find an open interval in any Lebesgue set of non zero measure, ...

This part of your solution is wrong: There exist Lebesgue sets of non-zero measure which contain no intervals. See for example 'fat cantor sets' or $[0,1] \setminus\mathbb{Q}$.
The first statement is also known as 'Steinhaus Theorem'. In order to prove the statement you can use the hint in the following way: We know that there exists a non-trivial interval $I=(a,b)$ with $4m(E \cap I) > 3m(I)$. Define $J= (-\delta,\delta)$ with $\delta = (b-a)/2$. If $(E \cap I) \cap (E\cap I +v) = \emptyset$ for some $|v| < \delta$, w.l.o.g. $v>0$ we find that
$$ \frac{3}{2} m(I)<2 \mu(E \cap I) = \mu(E \cap I \cup E \cap I +v) \leq \mu(I + [0,\delta))= m(I) +\delta< \frac{3}{2} m(I).$$
A contradiction! Thus, we have $(E \cap I) \cap (E\cap I +v) \neq \emptyset$ and hence $v \in E \cap I - E \cap I \subset E -E$. Since $|v| <\delta$ was arbitary, we see that $(-m(I)/2,m(I)/2) \subset E-E$.
Edit: I have decided to add a proof of the "previous exercise". For simplicity, let w.l.o.g $E \subset [0,1]$. We know for the Lebesgue measure (e.g. by the Caratheodory construction) that there exists an open set $U$ with $E \subset U$, finite measure and
$$\tag{1} m(E) \ge (1-\varepsilon) m(U).$$
Note that we have needed $m(E) >0$ at the last step. Write $U = \bigcup_{n=1}^\infty I_n$ with disjoint $I_n = (a_n,b_n)$, then (1) can be rewritten as
$$\tag{2}\sum_{n=1}^\infty m(E \cap I_n) \ge (1- \epsilon) \sum_{n=1}^\infty m(I_n). $$
However, (2) cannot hold if $m(E \cap I_n) \le (1-2 \epsilon) m(I_n)$ for all $n \in \mathbb{N}$. Thus, at least one natural number $n \in \mathbb{N}$ exists satisfying
$$m(E \cap I_n) \ge (1-2 \epsilon) m(I_n).$$
