$[a,b]\cap\mathbb{Q}$ are not compact as we may take a sequence converging to an irrational number (in $\mathbb{R}$ ) and no subsequence converges to a point in $\mathbb{Q}$ (sequential compactness is equivalent to compactness for metric spaces). Suppose some compact (and, therefore, closed) subset $S$ of $\mathbb{Q}$ contains an open subset of $\mathbb{Q}$ . Then it contains an interval $[a,b]\cap\mathbb{Q}$ . The interval is closed in $S$ and, therefore, compact. Contradiction. Therefore, there are no compact subsets of $\mathbb{Q}$ that contain any open subset. Hence, $\mathbb{Q}$ is not locally compact.