Solution verification: $m(A)>1$ implies $\exists x,y\in A$ s.t. $x-y\in\mathbb{N}.$

Let $A\subset\mathbb{R}$ with $m(A)>1$. Prove that there exists $x,y\in A$ such that $x-y$ is a positive integer. ($m$ is the Lebesgue measure.)

I think I have an interesting idea to solve this which I present below and would like to confirm if it is correct.

Fix $y\in A$. Then we need to show that there is a positive integer $n$ such that $x=y+n\in A$. i.e. $y\in A-n$. In words, my idea is to instead prove that there is a translate of $A$ that also contains $y$.

$$\int_{\mathbb{R}}\sum_{n=1}^\infty \chi_{A-n}(y)dy=\sum_{n=1}^\infty\int_{\mathbb{R}} \chi_{A-n}(y)dy= \sum_{n=1}^\infty m(A-n)>\sum_{n=1}^\infty1=\infty$$

This implies that $\sum_{n=1}^\infty\chi_{A-n}(y)>0$ which implies that that there exists a positive integer $n$ such that $y\in A-n$. Then $y=x-n$, for some $x\in A$ and thus $x-y=n$, as required.

Question: Is this correct? If not, can it be fixed so that my idea can still be applied?

Thanks a lot for your time.

HINT:

Say $A\subset [-d, d]$, place $(n+1)$ translates by $0$, $1$, $\ldots$, $n$ into a segment $[-d, d+n]$. For some $n$ large enough, we will have $$(n+1) \mu(A) > \mu([-d, d+n])$$ so two of the translates will intersect.