Here's the problem ($\lambda$ denotes the Lebesgue measure):
Let $A \subset R^2$ with $\lambda(A)>1$. Prove: $(A-A) \cap \mathbb{N}^2 \ne \emptyset$
After trying to prove it for sets $A \subset R$, I arrived at the following argument:
Let $A + n =\{x+n:x\in A\}$ for $n\in\mathbb{N}$. Assume by contradiction that the $A_n$'s are pairwise disjoint and consider the following set:
$$B_n = \frac{1}{n} \bigcup_{k=1}^n (k+A)$$
By translation invariance we get $\lambda(B_n) = \frac{1}{n} \sum_{k=1}^{n}\lambda(A)= \lambda(A)$.
Looking at the diameter we arrive at $diam(nB_n) = diam(A)+n \implies diam(B_n) = \frac{diam(A)}{n}+1$
Now comes the suspicious part...
Taking the limit as $n \to \infty$ we get $\lambda(A) = \lambda(B_n) \le diam(B_n) \to 1 <\lambda(A)$ contradiction.
I said suspicious because i don't see any formal way to justify taking the limit as $n \to \infty$
If you take $A$ long thin around the line $x+y=0$, you'll get no $P,Q\in A$ s.t. $P-Q=(k,l)$ with $k,l$ positive integers.
What you can prove is that $(A-A) \cap (\mathbb{Z}^2\setminus(0,0)) \ne \emptyset$. If you split $\mathbb R^2$ to (disjoint) squares $C_{m,n}=[m,m+1)\times[n,n+1)$ ($m,n\in\mathbb Z$) and set $A_{m,n}=(A\cap C_{m,n})-\{(m,n)\}\subset C_{0,0}$ then $\lambda(A)=\sum_{m,n}\lambda(A_{m,n})$. If $A-A$ contains no $(k,l)\neq(0,0)$ then $A_{m,n}$'s are disjoint, i.e. $\sum_{m,n}\lambda(A_{m,n})\leq\lambda(C_{0,0})=1$, a contradiction with $\lambda(A)>1$