A locally compact Hausdorff space is compactly generated

Question

I am having trouble showing one direction of the proof that a locally compact Hausdorff space is compactly generated.

Specifically, my question is how do I show that: if X is a locally compact Hausdorff space and A is a subset of X with the property that $$A\cap K$$ is closed for every compact K, then A is closed.

Answer 1

HINT: If $A$ is not closed, there is a point $x\in(\operatorname{cl}A)\setminus A$. Let $U$ be an open nbhd of $x$ with compact closure, and let $K=\operatorname{cl}U$. Now consider $K\cap A$ to get a contradiction.

Answer 2

Let $x$ be an element in the complementary of $A$, since $X$ is locally compact, there exists a neighborhood $U$ of $x$ which has a compact adherence $U'$, $A\cap U'$ is compact thus closed. It implies that $U-A\cap U'$ is open, thus the complementary of $A$ is open. done.