The difference is that if $X$ is compact, every collection of closed sets with the finite intersection property has a non-empty intersection; if $x$ is only countably compact, this is guaranteed only for countable collections of closed sets with the finite intersection property. In a countably compact space that is not compact, there will be some uncountable collection of closed sets that has the finite intersection property but also has empty intersection.
An example is the space $\omega_1$ of countable ordinals with the order topology. For each $\xi<\omega_1$ let $F_\xi=\{\alpha<\omega_1:\xi\le\alpha\}=[\xi,\omega_1)$, and let $\mathscr{F}=\{F_\xi:\xi<\omega_1\}$. $\mathscr{F}$ is a nested family: if $\xi<\eta<\omega_1$, then $F_\xi\supsetneqq F_\eta$. Thus, it certainly has the finite intersection property: if $\{F_{\xi_0},F_{\xi_1},\dots,F_{\xi_n}\}$ is any finite subcollection of $\mathscr{F}$, and $\xi_0<\xi_1<\ldots<\xi_n$, then $F_{\xi_0}\cap F_{\xi_1}\cap\ldots\cap F_{\xi_n}=F_{\xi_n}\ne\varnothing$. But $\bigcap\mathscr{F}=\varnothing$, because for each $\xi<\omega_1$ we have $\xi\notin F_{\xi+1}$. This space is a standard example of a countably compact space that it not compact.
Added: Note that neither of them says:
Given a collection of closed sets, when a finite number of them has a nonempty intersection, all of them have a nonempty intersection.
The finite intersection property is not that some finite number of the sets has non-empty intersection: it says that every finite subfamily has non-empty intersection. Consider, for instance, the sets $\{0,1\},\{1,2\}$, and $\{0,2\}$: every two of them have non-empty intersection, but the intersection of all three is empty. This little collection of sets does not have the finite intersection property.
Here is perhaps a better way to think of these results. In a compact space, if you have a collection $\mathscr{C}$ of closed sets whose intersection $\bigcap\mathscr{C}$ is empty, then some finite subcollection of $\mathscr{C}$ already has empty intersection: there is some positive integer $n$, and there are some $C_1,\dots,C_n\in\mathscr{C}$ such that $C_1\cap\ldots\cap C_n=\varnothing$. In a countably compact space something similar but weaker is true: if you have a countable collection $\mathscr{C}$ of closed sets whose intersection $\bigcap\mathscr{C}$ is empty, then some finite subcollection of $\mathscr{C}$ already has empty intersection. In a countably compact space you can’t in general say anything about uncountable collections of closed sets with empty intersection.
The more general fact is true. If $(C_\alpha)_{\alpha\in I}$ is a collection of closed subsets with finite intersection property, then all these subsets have a common point.
Assume $\cap_{\alpha\in I}C_\alpha=\varnothing$, then for open subsets of $C$ which we denote $U_\alpha=C\setminus C_\alpha$ we have $\cup_{\alpha\in I}U_\alpha=C$. Since $C$ is compact, we have finite collection $\{\alpha_1,\ldots,\alpha_n\}\subset I$ such that $C=U_{\alpha_1}\cup\ldots\cup U_{\alpha_n}$. Taking complements we get that $C_{\alpha_1}\cap\ldots C_{\alpha_n}=\varnothing$. Contradiction, so $\cap_{\alpha\in I}C_\alpha\neq\varnothing$
Best Answer
I think "closed intervals" means intervals of the form $[a,b]$. The statement is false if you include half-intervals. (Consider the collection $\{[1,\infty),[2,\infty),[3,\infty),\ldots\}$.)