First, to attempt to answer Gerry's question: I think that if $C$ is a closed set for which $f(C) = C$, then $\Omega(f, C)$ is supposed to be the set of non-wandering points of the restriction of $f$ to $C$, that is

$$\Omega(f, C) = \{x \in C: \forall U \in Nbhd(x) \forall n_0 \exists n \geq n_0 s.t. f^{(n)}(U) \cap U \neq \emptyset\}$$

where $Nbhd(x)$ consists of open neighborhoods of $x$ relative to $C$. I had to look this up myself and it's conceivable that I misunderstood something somewhere.

But I think we only need the fact that in a compact metric space, a strictly decreasing transfinite sequence of closed sets must terminate after countably many steps. That is to say, if $\gamma$ is an ordinal and $F: \gamma \to P(X)$ has the property: for $\alpha < \beta < \gamma$, there is a strict inclusion $F(\beta) \subsetneq F(\alpha)$ of closed subsets, then $\gamma$ is countable.

For this, we just need (1) compact metric spaces are separable, and (2) the conclusion holds for separable spaces. (1): a compact metric space has a countable dense subset (for each $n$, there is a finite cover by open balls of radius $1/n$, and the centers of these balls for $n$ ranging over natural numbers gives the countable dense subset). (2): Each closed subset in a separable space (with given countable dense set $D$) is characterized by the set of elements of $D$ is contains. Since each difference $F(\alpha) - F(\alpha+1)$ has nonempty interior and therefore contains at least one element of $D$, there can be at most countably many $\alpha < \gamma$.

**Edit:** Andreas Blass pointed out that this suggested proof is flawed, and also the easy fix in his comment below. The argument given for (1) shows that compact metric spaces have a countable base for the topology (second countability). Then replace (2) by "each closed subset in a second-countable space (with given countable base) is characterized by the open sets in the base it is disjoint from", and proceed similarly as above.