* Q is not locally compact, since for any positive irrational number ε, I(ε)=[-ε,ε] is not compact in Q.

Remember open set U in Q is a form U=V⋂Q where V⊆R is open subset. Thus any increasing open covering of I(ε), finitely many open subset of Q can NOT cover I(ε).

This concludes that the intersection of locally compact subspaces are not necessarily locally compact (this conclusion is based on the study of an introduced topological space explained in the link).

* In locally compact space X, the subset A is closed if and only if the intersections with all compact subset is compact. This can be proved as follows. Let K be an arbitrary compact subset of X, then (because of hausdorff X) K is closed and A as well by assumption. Then A⋂K is closed and again is compact. Reversely if A is open subset, then complement of A, denoted c(A) is closed. Fix a point x in the boundary of c(A), because X is locally compact, there is a compact neighbor of x, say K∋x. By definition of neighbor, K⋂A≠∅ and K⋂c(A)≠∅. Because K⋂c(A) is closed, K⋂A is open in K. Thus K⋂A cannot be closed in X, and therefore K⋂A cannot be compact (for any subset B⊆X, B can be closed in X necessarily when intersection with arbitrary closed subset A⊆X is closed due to the basic property that any closed subset is closed with operation of taking an intersection).