A subspace $A$ of $X$ is $\underline{\text{compactly closed}}$ if for any compact Hausdorff space $K$ and any continuous map $g: K \to X$, the pre-image $g^{-1}(A)$ is closed in $K$.
A topological space $X$ is $\underline{\text{weak Hausdorff}}$ if for any compact Hausdorff space $K$ and every continuous map $g:K \to X$, the image $g(K)$ is closed in $X$.
In Chapter 5 of May's algebraic topology, he mentioned that when $X$ is weak Hausdorff, $A\subset X$ is compactly closed implies that the intersection of $A$ with each compact subset of $X$ is closed.
I can see that $A\cap g(K)$ is closed in $X$, but I don't know how to generalize this to all compact subsets of $X$.
Best Answer
If $K \subseteq X$ is compact (which May uses to mean compact Hausdorff, as mentioned at the start of that paragraph (!)) then $i: K \to X$ (the embedding) is a continuous map from a compact space into $X$ so by weak Hausdorffness of $X$ we know that $i[K]=K$ is actually closed in $X$. And now, if $A$ is compactly closed, $i^{-1}[A]= A \cap K$ must be closed in $K$ and hence closed in $X$ too (closed in closed is closed).