Consider the map $f:\mathcal{P}\rightarrow [0,1)^2$ which maps $(x,y)\mapsto (\{x\},\{y\})$ (the fractional parts of $x$ and $y$) This map locally preserves area (it is composed of several translations), so $|f(\mathcal{P})|\leq |\mathcal{P}|$. Assume there are no points in $\mathcal{P}$ differing by an integer vector, i.e. $f$ is injective, and so we have equality case. Since $f(\mathcal{P})\subset [0,1)^2$, $|\mathcal{P}|=|f(\mathcal{P})|\leq 1$, but this is clearly a contradiction.