How is compactness related to countable saturation? By Cantor's intersection theorem every decreasing nested sequence of nonempty compact sets has a common point. 
A superficially similar result holds that every decreasing nested sequence of nonempty internal sets in an ultrapower model of a hyperreal field ${}^\ast\mathbb{R}$ has a common point, a property known as countable saturation. 
Is such a resemblance more than superficial?
 A: As Todd Trimble pointed out in the comments, the use of the term "compactness" in the Compactness Theorem does refer to the topological notion: the Stone space of the Lindenbaum algebra of a theory is compact. For saturation, the relevant notion are the spaces of complete $n$-types. These are naturally compact spaces. If $A_0 \supseteq A_1 \supseteq A_2 \supseteq \cdots$ are nonempty internal sets, then $\{x \in A_0,x \in A_1,x \in A_2,\ldots\}$ extends to a complete $1$-type $p(x)$ by compactness. A countably saturated model is one which realizes every complete type over a countable parameter set (e.g., $A_0,A_1,A_2,\ldots$). Therefore, a countably saturated model of non-standard analysis must realize the type $p(x)$. Because models of non-standard analysis satisfy comprehension, one can do without the syntactic notion of types and simply work with sets. This way, countable saturation for such models is equivalent to saying that the intersection of a countable family of internal sets is nonempty.
A: It turns out that there is a direct relationship between Cantor's lemma for nested compact sets $K_n$, on the one hand, and nested sequences of internal sets, on the other. Namely, the latter can be viewed as a special case of the former as follows.
The decreasing nested sequence of internal sets, $\langle{}^\ast\!K_n\colon n\in\mathbb N\rangle$, has a common point $x$ by saturation. But for a compact set $K_n$, every point of ${}^\ast\!K_n$ is nearstandard, i.e., infinitely close to a point of $K_n$ (in other words, every point is in the halo of a standard point of $K_n$). In particular, $st(x)\in K_n$ for all $n$, as required. 
