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.
A topological space $X$ is compact if: for every covering of $X$ with open sets there exists a finite subcovering. You want to prove that this property is equivalent to: for every family of closed sets such that every finite subfamily has nonempty intersection then the intersection of the whole family was nonempty.
The equivalence is very simple: to pass from one statement to the other you have just to pass to the complementary of sets.
So: open becomes closed, union becomes intersection, covering becomes empty intersection. With this translation you get the desired equivalence.
Formally: Compactness means that for every family $\mathcal R$ of open sets:
$$
\bigcup \mathcal R = X \Longrightarrow \exists \text{finite}\ \mathcal R_0 \subset \mathcal R \colon \bigcup \mathcal R_0 = X,
$$
while the other property is that for every family $\mathcal F$ of closed sets:
$$
\bigcap \mathcal F \neq \emptyset \Longleftarrow \forall\text{finite}\ \mathcal F_0\subset \mathcal F \colon \bigcap \mathcal F_0 \neq \emptyset.
$$
You can reverse the implication by negating both sides. Hence passing to the complementary (as said before) you get the equivalence.
Best Answer
There is no problem, because you don’t have to use the compactness of the collection to prove the theorem.
In fact if $X$ is compact and if $\{C_\alpha\}$ is a collection of closed sets with the FIP property, then by contradiction if $\cap C_\alpha= \emptyset$ then $\cup C_\alpha^c=X$ and so $\{C_\alpha^c\}$ is an open cover of $X$ that does not admits a finite subcover. Thus $X$ cannot be compact.
You can prove the inverse one in a similar way.
Now the definition of FIP property is usually done with a collection of ‘sets’. The theorem holds also for only a collection of closed subsets and the definition of your book it makes sense because if $X$ is compact then each closed subset is also compact and so each collection of closed subsets of $X$ has to be a collection of compact subsets of $X$.