Equivalent criteria for being a compactly closed set A space $X$ is said to be "weak Hausdorff" (WH) if $g(K)$ is closed in $X$ for every map $g:K\to X$ from a compact Hausdorff space $K$ into $X$. A subspace $A$ of $X$ is said to be "compactly closed" if $g^{-1}(A)$ is closed in $K$ for any (continuous) map $g:K\to X$ from a compact Hausdorff space $K$ into $X$.
Let us now assume that $X$ is WH. In May's algebraic topology book, it says that $A\subseteq X$ is compactly closed $\iff$ the intersection of $A$ with each compact set of $X$ is closed. I am having trouble proving this, however. The $\Leftarrow$ implication is relatively easy to prove, but I'm having trouble proving the $\Rightarrow$ implication. I would appreciate any hints for this problem!
 A: May has the convention that "compact" includes "Hausdorff". With your terminology May says that $A\subset X$ is compactly closed $\iff$ the intersection of $A$ with each compact Hausdorff subset of $X$ is closed.
$\Rightarrow$
Let $C \subset X$ be compact Hausdorff. Since $A$ is compactly closed, the preimage $id_C^{-1}(A) = A \cap C$ is closed in $C$.
$\Leftarrow$
Let $K$ be compact Hausdorff and $g : K \to X$ be continuous. We claim that $g(K)$ is compact Hausdorff. Compactness is trivial. Let $x_1, x_2 \in g(K)$ with $x_1 \ne x_2$. The sets $\{x_i\}$ are closed (being the images of $\{x_i\} \hookrightarrow X$), thus the subsets $g^{-1}(x_i)  \subset K$ are closed and disjoint. Since $K$ is normal, we find open neighborhoods $U_i$ of $g^{-1}(x_i)$ in $K$ such that $U_1 \cap U_2 = \emptyset$. The sets $g(K \setminus U_i)$ are closed in $X$, thus closed in $g(K)$. Hence the sets $V_i = g(K) \setminus g(K \setminus U_i)$ are open in g(K). It is easy to see that $x_i \in V_i$ and $V_1 \cap V_2 = \emptyset$. Hence $g(K)$ is Hausdorff.
We conclude that $A \cap g(K)$ is closed in $g(K)$, thus $g^{-1}(A)$ is closed in $K$.
