Is $\mathbb R$ with cocountable topology star-$K$-compact?

Question

1. A space $X$ is said to be starcompact if for every open cover $\mathcal U$ of $X$ there exists a finite subset $\mathcal V\subseteq\mathcal U$ such that $St(\cup\mathcal V,\mathcal U)=X$.

2. A space $X$ is said to be star-$K$-compact if for every open cover $\mathcal U$ of $X$ there exists a compact subset $K$ of $X$ such that $St(K,\mathcal U)=X$.

It is immediate that $\mathbb R$ with cocountable topology is starcompact. But we don't know whether $\mathbb R$ with cocountable topology is star-$K$-compact. It is clear from the definitions that every star-$K$-compact space is starcompact.

Answer 1

Yes, $\mathbb R$ with the cocountable topology is star-$K$-compact. If $\mathcal U$ is an open cover of $\mathbb R$ then, as any open set misses only countably many points, there is a countable $\mathcal U'\subseteq\mathcal U$ still covering $\mathbb R$. Now $\mathbb R$ is not the countable union of countable sets so that $\bigcap \mathcal U'\neq\emptyset$. It follows that there is a point $x$ with $\mathbb R=St(\{x\},\mathcal U')=St(\{x\},\mathcal U)$.