= Solution
The <Minkowski convex body theorem> states that a centrally symmetric convex measurable set in $\mathbb R^d$ of volume greater than $2^d\operatorname{covol}(\Lambda)$ contains a nonzero point of the full lattice $\Lambda$. For a compact convex body, the weak inequality also suffices: apply the strict version to $(1+\varepsilon)C$, let $\varepsilon\downarrow0$, 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 $\Lambda=\mathbb Z^2$ to
$$
C_Q=\{(u,v):|u|\le Q,\ |xu-v|\le Q^{-1}\}.
$$
The 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 $n\ne0$. It follows that
$$
\boxed{\left|x-\frac mn\right|\le\frac1{Q|n|}\le\frac1{n^2}.}
$$
If $x$ 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 $j\ge1$ have zero error and are infinitely many distinct pairs. This proves the assertion for every real $x$.
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