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.
This theorem holds also for topological spaces that are not metric, so using Cauchy sequences might not be the most suitable approach.
Let $X$ be a topological space and assume that the intersection of any family of closed subsets having the finite intersection property is non-empty. Let $\{O_i\}_{i\in I}$ be some open cover of $X$ and assume that it has no finite subcover. Hence for each finite $F\subseteq I$ there is some $x\in X$ such that $x\notin \bigcup_{i\in F}O_i$. Equivalently, for each finite $F\subseteq I$ there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$. Hence the family $\{X\setminus O_i\}_{i\in I}$ is a family of closed subsets with the finite intersection property. It follows that there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$, i.e., $x\notin\bigcup_{i\in I}O_i$, contradicting that $\{O_i\}_{i\in I}$ is a cover of $X$. We conclude that $\{O_i\}_{i\in I}$ must have a finite subcover, so $X$ is compact.
Conversely, assume that $X$ is compact and let $\{C_i\}_{i\in I}$ be some family of closed subsets with the finite intersection property. Assume that $\bigcap_{i\in I}C_i$ is empty. Then $\{X\setminus C_i\}_{i\in I}$ is an open cover of $X$. By compactness, it has a finite subcover, so $\bigcup_{i\in F}X\setminus C_i=X$ for some finite $F\subseteq I$. But then $\bigcap_{i\in F}C_i=\emptyset$, contradicting that $\{C_i\}_{i\in I}$ has the finite intersection property. We conclude that $\bigcap_{i\in I}C_i\neq\emptyset$.
Best Answer
This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $x\in U$ and $y\leq x$ implies $y\in U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $i\in (1/2,1]$, let $C_i=[0,1/2)\cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.