Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):
Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and
$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$
is an open cover of $\,X\,$ and thus there exists a finite subcover of it:
$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$
The family $\,\{V_i\}\,$ has the FIP.
(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:
$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$
By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then
$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$
$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.
Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...
Best Answer
Let say $\bigcap_{n \in \mathbb{N}} K_n=\emptyset$ . Then $(\bigcap_{n \in \mathbb{N}} K_n)^c=\emptyset^c$. It means $\bigcup_{n \in \mathbb{N}} K_n^c=E$, so the familly $\{K_n^c:n\in \mathbb{N} \}$ is open cover of $E$. Since $E$ is compact we have a finite subcover of the familly. It means there exists a finite subset $S$ of $ \mathbb{N}$ such that $\bigcup_{n \in S} K_n^c=E$. So $(\bigcup_{n \in S} K_n^c)^c=E^c$, it means $\bigcap_{n \in S} K_n=\emptyset$. Buy using decreasing property of the sequence, we have the smallest one of the familly $\{K_n:n\in S\}$. Say it is $K_{n_0}$ and so $K_{n_0}=\bigcap_{n \in S} K_n=\emptyset$. But all the $K_n$'s are non-empty. This is contradiction.
For the third one. $K_{\infty}$ is compact since it is closed and subset of a compact set say $K_n$