Let $A,B$ be two measurable set with positive measure. Show that $A-B$ contains rational number

As you have observed, the point is to reduce to the case of $A = B$, which we can do by translation: consider the family of translates $x + A$ of $A$ (with $x \in \mathbb{R}$). Now define the function $f(x) := \mu((x + A) \cap B)$ (which is everywhere nonnegative). Observe that by Tonelli's theorem we have \begin{align*} \int_{\mathbb{R}} f(x) \ d x &= \int_{\mathbb{R}} \mu((x + A) \cap B) \ d x \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{1}_{x + A}(y) \mathbb{1}_{B}(y) \ d y \ d x \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{1}_{A}(y - x) \ d x \ \mathbb{1}_{B}(y) \ d y\\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{1}_{A}(x) \ d x \ \mathbb{1}_{B}(y) \ d y\\ &= \int_{\mathbb{R}} \mu(A) \ \mathbb{1}_{B}(y) \ d y\\ &= \mu(A) \mu(B). \end{align*} If $A$ and $B$ both have positive measure, then this integral is therefore strictly positive.

We immediately conclude that there is some real $x_0 \in \mathbb{R}$ such that $f(x_0)$ is nonzero, i.e. such that $\mu((x_0 + A) \cap B) > 0$. We'd like to be able to set $C = (x_0 + A) \cap B$ and now appeal to the Steinhaus theorem (just for $C$), but this doesn't work because $x_0$ might be irrational. But we can easily fix this: note that as we used above, $f$ is equivalently defined by the integral $$ f(x) = \int_{\mathbb{R}} \mathbb{1}_{x + A}(y) \mathbb{1}_B(y) \ dy = \int_{\mathbb{R}} \mathbb{1}_{A}(y - x) \mathbb{1}_B(y) \ dy, $$ which is (up to a reflection of $A$ about $0$) the convolution of the two functions $\mathbb{1}_A$ and $\mathbb{1}_B$, and convolutions are easily shown to be continuous (e.g. see this question).

We can now fix some $x_0 \in \mathbb{R}$ such that $f(x) > 0$, and choose a sequence of rationals $(q_n)$ converging to $x_0$. Then since we now know that $f(x)$ is continuous in particular the sequence $f(q_n)$ converges to $f(x_0)$, so there must be some $k \in \mathbb{N}$ such that $f(q_k) > 0$. In other words $\mu((q_k + A) \cap B) > 0$, so we can now appeal to the Steinhaus theorem with $C = (q_k + A) \cap B$.

A proof of this can be obtained by slightly modifying Steinhaus's Theorem Proof, and particular the one appearing here.

Step 1. Assume that $A$ and $B$ have both finite positive measure (otherwise, if for example $m(A)=\infty$, replace it with $\tilde A=A\cap [-M,M]$, for $M$ sufficiently large.)

Step 2. For $\varepsilon>0$, there exist $U$ and $V$ open sets such that $A\subset U$ and $B\subset V$ and $$ m(A)+\varepsilon>m(U) \quad\text{and}\quad m(B)+\varepsilon>m(V). $$

Step 3. For $\varepsilon>2/3$, choose intervals $I\subset U$ and $J\subset V$, such that for $E=A\cap I$ and $F=B\cap J$, $$ m(E)>\frac{2}{3}m(I) \quad\text{and}\quad m(F)>\frac{2}{3}m(J). $$

Step 4. Fix $I$ and $J$ so that they have the same size. If for example $m(I)<m(J)$, find a subinterval $J'\subset J$, so that $m(J')=m(I)$ and $m(E')>\frac{2}{3}m(J')$, where $E'=J'\cap B$.

Step 5. Now that $m(I)=m(J)$, there exists an $s$, such that $I=s+J$ and hence $$ E,\,F+s\subset I \quad\text{and}\quad m(E),\,m(F+s)> \frac{2}{3}m(I), $$ and as in the proof of Steinhaus's Theorem, we will conclude that $$ (-\delta,\delta)\subset (E-s)-F $$ for some $\delta>\frac{1}{6}$. Consequently $$ (s-\delta,s+\delta) \subset E-F\subset A-B. $$