The rationals are not locally compact

Suppose \(\mathbb Q\) were locally compact at \(q\). Then some neighborhood \(V\) of \(q\) in \(\mathbb Q\) would have compact closure in \(\mathbb Q\). Choose real numbers \(a<b\) with \[ q\in(a,b)\cap\mathbb Q\subset V. \] Pick an irrational \(r\in(a,b)\). Choose a sequence of rationals \(q_n\in(a,b)\) converging in \(\mathbb R\) to \(r\). The sequence lies in \(\overline V^{\mathbb Q}\). If that closure were compact, some subsequence would converge in \(\mathbb Q\) to a point \(s\in\mathbb Q\). But convergence in the subspace \(\mathbb Q\subset\mathbb R\) implies convergence in \(\mathbb R\), and uniqueness of real limits gives \(s=r\), contradiction. Hence \(\mathbb Q\) is not locally compact.