Recently, I read a paper entitled "Hyper-expansive homeomorphisms".
Definition: A homeomorphism $f:X\to X$ is hyper-expansive if there is a $c>0$ such that for every closed sets $A\neq B$, there is $n\in \mathbb{Z}$ with $d_H(f^n(A), f^n(B))>c$, where $d_H$ is Hausdorff distance.
The author claimed that hyper-expansiveness of $f:X\to X$ implies that $f$ has a finite number of orbits, hence $X$ is countable. In this direction, he showed that if $K$ is a minimal set, then $K=\{p\}$ is a fixed point. This implies that for every $x\in X$, $\omega_f(x)=\{y\}$, a fixed point.
Note that hyper-expansivity implies expansivity, hence the set of fixed points is finite.
But there is some statement in its proof that is not clear for me:
Let $f:X\to X$ be hyper-expansive homeomorphism with hyper-expansive constant $c>0$.
1) He claimed that for every $\epsilon>0$ there is $n\in \mathbb{N}$ such that if $x\notin B_\epsilon(\Omega(f))$, then $f^j(x)\in B_\epsilon(\Omega(f))$ for $|j|>n$(???)
2) Also he claimed that $f$ has finite number of orbits, because if it has infinite number of orbits, then $X-B_\epsilon(\Omega(f))$ is an infinite compact set. There are $x,y \notin B_\epsilon(\Omega(f))$ and $p, q\in \Omega(f)$ such that $\omega(x)=\omega(y)= \{p\}$ and $\alpha(x)=\alpha(y)=\{q\}$. Then if $d(x, y)$ is small, this two points contradicts the expansiveness of $f$(???)
Would you please help me to know (1), (2).