 Solution

The Minkowski convex body theorem states that a centrally symmetric convex measurable set in ${\mathbb{R}}^d$ of volume greater than $2^d\mathrm{covol}(\Lambda )$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\epsilon )C$, let $\epsilon \downarrow 0$, and choose a constant subsequence of the resulting lattice points in the bounded set $2C$. Its nonzero value belongs to $C$ by closedness.

For $Q>1$, apply this borderline version with $\Lambda ={\mathbb{Z}}^2$ toThe linear transformation $(u,v)\mapsto (u,xu-v)$ has determinant of absolute value one, so this parallelogram has area four. A nonzero lattice point $(n,m)$ satisfies $|n|\le Q$ and $|nx-m|\le Q^{-1}$. If $n=0$, then $|m|<1$, giving the forbidden zero point; hence $n\ne 0$. It follows thatIf $x$ is irrational, these pairs cannot remain in a finite set as $Q\to \infty$: the positive errors $|nx-m|$ of any such finite set have a positive minimum, contradicting their bound $Q^{-1}$. If $x=p/q$ is rational in lowest terms with $q>0$, the pairs $(n,m)=(jq,jp)$ for $j\ge 1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.