If $E$ has positive measure, then prove that there exists $h \in \mathbb{R}$ such that $|(E+h) \cap E| > 0$.

It suffices to prove the statement for bounded $E$. So suppose $E$ is bounded.

Note $\int\int 1_E(y)1_E(y-h)dydh = \int 1_E(y)\int 1_E(y-h)dhdy = \int 1_E(y)|E|dy = |E|^2 > 0$.

So, there exists $h$ with $\int 1_E(y) 1_E(y-h)dy > 0$. But this integral is $|E\cap (h+E)|$.

If $E$ has positive measure then there exists an interval, say $I$ such that $m(E\cup I)>\frac{2}{3}m(I)$. (Of course much more can be said about points of density of $E$ but we do not need it here). Fix this interval $I$. Now, if $E-E$ does not contain a neighborhood of $0$ then there is a decreasing to zero sequence, say $x_{n}$ such that $E+x_n$ is disjoint with $E$. In particular, the set $I\cap E$ would be disjoint with $(I+x_n)\cap (E+x_n)$. However, $E+x_n$ is just a shift of $E$ so $m((I+x_n)\cap(E+x_n))> \frac{2}{3}m(I+x_n)$. Clearly, $m(I+x_n)=m(I)$. Now, this is not possible since for $x_n$ sufficiently small (i.e. $x_n< \frac{1}{6}m(I)$ ) we would have two disjoint sets contained in a small perturbation of $I$ ( add and subtract $x_n$ from I boundaries) each having measure larger than $\frac{2}{3}$ of $I$.
So, $E-E$ contains a neighborhood of $0$ which is the same as to say that $m(E\cap(E+h)) > 0$ for $h$ small enough.