We have a topological space $A$. I want to prove that it's locally compact subspace $B$ always contains non-empty open subset if $B$ is not nowhere-dense.
I am not even sure is it true or not, may be it is possible to construct a counterexample.. But the statement in the title seems to be right to me.There are no conditions on $A$, so it is not even Hausdorff.
I've tried to achieve a contradiction assuming that $B$ does not contain any open subset (so all points of B are boundary points) and it's closure contains at least one.
But I have no idea how to use locally compactness.