Theorem (Steinhaus): If $E \subseteq \mathbb{R}$ is a measurable set with positive Lebesgue measure, then the set $$ E - E = \{e_1 - e_2 : e_1, e_2 \in E\} $$ contains an open interval around the origin.
Let $m$ denote the lebesgue measure. It suffices to prove the case where $m(E) < \infty$, since $m$ is a semifinite measure. There exists an interval $(a,b)$ where $a,b\neq \pm \infty$, such that $m((a,b)\cap E) > \frac{3}{4} m((a,b))$ (by a well-know theorem? Chapter 1 exercise 30 in folland). Let $I = [a,b)$. It is clear that $I$ satisfies the same inequality since it differs $(a,b)$ by a measure of $0$. We we will show that $(-\frac{1}{10}m(I),\frac{1}{10}m(I)) \subseteq E-E$. Suppose for sake of contradiction that $\exists k \in (-\frac{1}{10}m(I),\frac{1}{10}m(I))$ such that it is not in $E-E$. We assume $k>0$, since $E-E$ is clearly symmetric. Consider the intervals $I_0 = [a,a+k), I_1 = [a+k,a+2k),\dots, I_n = [a+nk,a+(n+1)k)$ where $a+nk<b\leq a+(n+1)k.$ Basically $\{I_j\}$ is the minimal half interval cover of $[a,b)$ each of length $k$. Then $$m(E\cap I)\leq m(\bigcup_{i=0}^n E \cap I_i)=\sum_{i=0}^nm(E\cap I_i),$$ since $I \subseteq \cup I_i$ and $\{I_i\}$ are pairwise disjoint. Assume that $n+1$ is even. The odd case is similar with some trivial details in the boundary case. Let $\omega_0 = m(E\cap I_0), \omega_1 = m(E\cap I_1)\omega_2 = m(E\cap I_2), \dots ,\omega_n = m(E\cap I_n)$. Notice that $E\cap I_{2t+1} \subseteq I_{2t+1} \setminus (E \cap I_{2t} + k)$, since if $x \in E\cap I_{2t+1}$ and $x \in (E \cap I_{2t} + k)$, then $x=y+k$ for some $y \in E \cap I_{2t}$, so that $x-y = k$, a contradiction to $k\notin E-E$. Thus $m(E\cap I_{2t+1}) \leq m(I_{2t+1} \setminus (E \cap I_{2t} + k)) = m(I_{2t+1}) - m(E \cap I_{2t} + k)$ . By translation invariance of lebesgue measure and the fact that each interval is of length $k$, it is also equal to $k-\omega_{2t}$. Thus, $\omega_{2t}+\omega_{2t+1} \leq k$. So $$\sum_{i=0}^nm(E\cap I_i)\leq \frac{n+1}{2}k \leq \frac{m(I)/k+1}{2}\cdot k= \frac{m(I)}{2}+\frac{1}{2}k<\frac{m(I)}{2}+\frac{m(I)}{10}<\frac{3}{4}m(I),$$ So $m(E\cap I) < \frac{3}{4}m(I)$. But $m(E\cap I) > \frac{3}{4} m(I)$ by our choice of $I$, a contradiction.
Here is how I remember the proof. Since $0<m(E)<\infty$, there is an open set $U$ containing $E$ such that $\displaystyle\frac{m(E)}{m(U)}>0.99$. Argue that if we express $U$ as the disjoint union of open intervals, then for some interval $(a,b)$, $E$ occupies at least $99\%$ of it, i.e., $\displaystyle\frac{m((a,b)\cap E)}{m((a,b))}\geq 0.99$.
Now we claim that $E-E$ contains the interval $(-\frac{b-a}{2},\frac{b-a}{2})$. Suppose not, so there exists $-\frac{b-a}{2}<z<\frac{b-a}{2}$ such that for any $x,y\in E$, $z\neq x-y$. Now we rephrase this a little bit:
$$\forall x,y\in E\ z\neq x-y$$ $$\forall x,y\in E\ x\neq y+z$$ $$\forall x\in E\ x\notin E+z$$ $$E\cap(E+z)=\emptyset$$
Since $E$ occupies $99\%$ of $(a,b)$, we know $E+z$ occupies $99\%$ of $(a+z,b+z)$, but these two intervals have a pretty large overlap (since we are just shifting $(a,b)$ by a number less than its radius), which is "obviously" a contradiction. More precisely, let $I$ be the intersection of these two intervals, then $m(I)>\frac{b-a}{2}$, so $E$ occupies at least $98\%$ of $I$; otherwise it cannot occupy $99\%$ of $(a,b)$ even if it contains the whole interval $(a,b)\setminus I$. Similarly $E+z$ occupies at least $98\%$ of $I$, which is a contradiction since they are disjoint.