An example of subset $A$ such that $A \cap K$ is open in $K$ for each compact set $K$, but $A$ is not open.

Question

Let $X$ be a topological space. For any $A \subseteq X$, consider two possible conditions on $A$:

1) $A$ is open in $X$;

2) $A \cap K$ is open in $K$, for each compact set $K \subseteq X$.

Then $(1)$ implies $(2)$. but, necessarily $(2)$ doesn't imply $(1)$. Give an example of $A$ such that $A$ satisfies in $(2)$ but $A$ is not open in $X$.

Answer 1

Take $X$ to be an uncountable set with the cocountable topology (a set is open if and only if its complement is countable). You can verify that the compact subsets of $X$ are finite sets with the discrete topology.

Hence, take any $A\subseteq X$ that is not open. Then, for any compact $K\subseteq X$, $A\cap K$ is open in $K$.

Answer 2

A space $X$ is anticompact if and only if the only compact subsets of $X$ are the finite subsets. Let $X$ be $T_1$, anticompact, and not discrete. Since $X$ is not discrete, it has a non-empty subset $A$ that is not open. $X$ is $T_1$, so the relative topology on each finite subset of $X$ is discrete, and therefore $A\cap K$ is open in $K$ for each compact $K\subseteq X$. Thus, any such $X$ provides an example.

Such spaces can even be Hausdorff: this answer by bof contains three of them, and I gave another in this answer. Three of those four examples are even zero-dimensional and normal.