Spaces in which "$A \cap K$ is closed for all compact $K$" implies "$A$ is closed."

Let $Q$ be the space of non-negative rational numbers, $N$ the natural numbers including $0$, and $q:Q\to Y=Q/N$ the quotient map identifying $N$ to a single point. Note that $Y$ is a Hausdorff normal space since $Q$ is such a space and $q$ is a closed map.
Now look at $Y\times Q$. This space is Hausdorff. We will show that it's not compactly generated.
Let us call the image of $(n,n+1)$ under $q$ the cell $e_n$. Note that a compact subset $C$ of $Y$ can only intersect finitely many cells. Otherwise, we could construct a subset $S\subseteq C$ by picking a point $s_n$ from every non-empty intersection $e_n\cap C$. Every subset $T$ of $S$ then had a closed preimage under $q$, thus would be closed, hence $S$ would be an infinite subset of $C$ without a limit point. This also means that a compact set in $Y\times Q$ is contained in a finite number of "cylinders" $\overline{e_n}\times Q$.
We now construct a non-closed set $A$ in $Y\times Q$ such that $A\cap K$ is closed for every compact $K$.
Let $h=q\times\mathbf 1_Q$ and let
$$ A' = \left\{(a,b)\in Q^2\ \middle|\ \frac\pi{\lfloor a+1\rfloor} < b < \frac\pi{\lfloor a+1\rfloor} +\min(a-\lfloor a\rfloor,\lceil a\rceil-a)\right\}. $$ This set is closed and the preimage of $A=h(A')$. However, $A$ is not closed as $(0,0)$ is its limit point. If $K$ is compact, then $A\cap K=h(A'\cap h^{-1}(K))$. The set $A'\cap h^{-1}(K)$ is closed and contained in some $[0,m]\times Q$, and that means $A\cap K$ is the image of a closed set under the restriction $h|_{[0,m]\times Q}$. This map is closed, being the product of the perfect map $q|_{[0,m]}$ with the identity on $Q$ (where perfect is the terminology for a map which is closed and has compact fibers). So we see that $A \cap K$ is closed, even though $A$ is not closed.