Counterexample on weaker version of result about compact sets

Question

The following is a very well known theorem:

Let X be a metric space. $K \subset X$ is compact iff every collection $ \{ F_j \}_{j\in A}$ of closed sets with the finite intersection property in K (that is, if $B \subset A $ is finite then $\cap\{ F_j \}_{j \in B} \cap K \neq \emptyset$) satisfies $ \cap \{ F_j \}_{j\in A} \cap K \neq \emptyset $.

Does anyone have an example to prove that this result is false if K is not compact, but only closed?

Answer 1

How about using $K=[1,\infty)\subseteq\mathbb{R}$ and for $j\in\mathbb{N}, F_j=[j,\infty)$