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.
If $Q_k$ has only finitely many points, then we have that $|Q_k|\geq|Q_{k+1}|\geq\ldots$, this is a decreasing sequence of positive integers, so it must stabilize. This means that for some $n$, every $m>n$ satisfies $Q_m=Q_n$, so clearly the intersection is non-empty, since it is equal to $Q_n$ as well.
The problem when each $Q_k$ is infinite is that infinite sets allow us to push things around, and since removing points from infinite sets might not change it is possible that at every step of the way $|Q_k|=|Q_n|$, but their intersection is suddenly empty after all. Think about the intersection of $A_n=\{k\in\Bbb N\mid k>n\}$ -- all these sets are infinite, but their intersection is empty!
Best Answer
A closed an convex set $C$ in a Banach space is also weakly closed: Suppose $x \not\in C$, then by Hahn-Banach (the seperating hyperplane theorem, I suppose) there is a functional $x^* \in X^*$ and a $\alpha \in \mathbb R$ such that $\sup_{y\in C}\Re x^*(y) \le \alpha$, $\Re x^*(x) > \alpha$. Now $\{\Re x^* > \alpha\}$ is a weakly neighbourhood of $x$ disjoint from $C$, so $X\setminus C$ is weakly open, hence $C$ is weakly closed.
For the non-empty intersection part, we use the weak compactness: As the $C_n$ are decreasing and not empty, finitely many of them have non-empty intersection, as they are compact, $\bigcap_n C_n \ne \emptyset$.