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**Aradesh** I've worked out that the book was right after all - though the reasoning is not obvious (to me). The claim is: Let $\displaystyle G$ be a Hausdorff space and $\displaystyle H \subset G$ is a locally compact subspace. If $\displaystyle x \in H$ then there is a closed neighbourhood $\displaystyle C \subset G$ of $\displaystyle x$ such that $\displaystyle H\cap C$ is a compact neighbourhood of $\displaystyle x$ in the subspace topology of $\displaystyle H$.

Proof of claim: As $\displaystyle H$ is locally compact we have a compact n'hood $\displaystyle K\subset H$ of $\displaystyle x$. This means that $\displaystyle K$ is compact in both $\displaystyle H$ and $\displaystyle G$ and there is an open $\displaystyle U \subset G$ containing $\displaystyle x$ such that $\displaystyle H \cap U \subset K$. Now $\displaystyle H$ is locally compact and Hausdorff and this means that it is a regular space. So since $\displaystyle H \cap U$ is an open n'hood of $\displaystyle x$ in the regular space $\displaystyle H$, there is an open n'hood $\displaystyle V \subset G$ of $\displaystyle x$ such that $\displaystyle H \cap \overline V \subset H \cap U$. It is easy to check that $\displaystyle C = K \cup \overline V \subset G$ is our desired closed n'hood of $\displaystyle x$.

Phew.