Compactness of $X$ and $ \cap \overline{U_n} = \emptyset$ The following is from the Munkres' Topology textbook : 

I have 3 questions regarding the underlined part of the text which I can't prove/disprove them:
1- Why $X$ must be compact so the intersection $\cap \overline{U_n}$ be nonempty?
2- Is there a non-compact space such that $ \cap \overline{U_n} = \emptyset$? How?
3- Does the converse hold, i.e. if the intersection $\cap \overline{U_n}$ is nonempty then $X$ is compact? How?
Sipmle detailed explanation would be much appreciated.    
 A: If $F_n \supseteq F_{n+1} \supseteq F_{n+2} \ldots$ is a decreasing family of non-empty closed sets in a compact $X$, then if $\cap_n F_n = \emptyset$, then $X\setminus F_n$ is an open cover of $X$ (if $x \in X$ then $x \notin F_k$ for some $k$, as we assumed the intersection is empty, and so $x \in X\setminus F_k$.
So by compactness $U_n$ has a finite subcover and as $U_n \subseteq U_{n+1} \subseteq U_{n+2} \ldots$ is increasing, the largest indexed $U_k$ in this subcover should by itself equal $X$, which cannot be as $F_k$ is non-empty (so its elements are not covered by $U_k$..) Contradiction , so $\cap_n F_n \neq \emptyset$.
We need some condition on $X$ here as otherwise we can take $F_n = \{n, n+1, n+2, \ldots \}$ in $X= \mathbb{N}$ in the discrete topology which has empty intersection, but $X$ is quite non-compact.
This property of $X$ for all decreasing families of non-empty closed sets, does not characterise compactness of $X$ but "countable compactness" of $X$ (every countable open cover of $X$ has a finite subcover). The proof is a lot like I gave above. 
