TITLE: Every locally compact space is compactly generated
QUESTION [8 upvotes]: I am using the following definitions (from Wikipedia):

A space $X$ is locally compact if every $x \in X$ has a compact neighborhood;
A space $X$ is compactly generated if a subset $A \subseteq X$ is closed if and only if $A \cap K \subseteq K$ is closed in every compact subset $K \subseteq X.$

According to Wikipedia's article on compactly generated spaces,

Every locally compact space is compactly generated.

I proved this claim for locally compact Hausdorff spaces, where compact subsets are also closed. My proof doesn't work in a non-Hausdorff case. I would like help with this case.

REPLY [3 votes]: Suppose $X$ is locally compact. Let $A$ be a subset of $X$ having the property that for every compact subspace $K$ of $X$, the intersection $K\cap A$ is closed $K$. To show that $A$ is closed in $X$, we show that every point in $X\setminus A$ has a neighborhood in $X$ disjoint from $A$.
Let $x\in X\setminus A$. Choose a compact subspace $K$ of $X$ that contains a neighborhood $U$ of $x$ in $X$. By hypothesis, $A_K:=K\setminus A$ is open in $K$; thus $U\cap A_K$ is open in $U$ and hence in $X$. This is the desired neighborhood of $x$.