Let $f:X\to X$ be a homeomorphism on compact metric space $(X, d)$ and $K$ be a minimal set, i.e. a compact $f$-invariant set that for all $x\in K$, the orbit $O(x)$ is dense in $K$.
In a paper, author claim that for every $\epsilon>0$ there is $n\geq 0$ such that for all $x\in X$, the set $O_n(x)=\{f^{i}(x)\}_{i=-n}^n$ is $\epsilon$-dense i.e. for all $y\in K$ there is $j\in \{-n, -n+1, \ldots, n\}$ such that $d(y, f^j(x))<\epsilon$.
Proof of it, is not clear for me. Please help me to know it.
Since $\{f^i(x)\}_{i= -\infty}^{\infty}$ is dense, the set of all balls $B_{\epsilon}(f^i(x))$ centred at $f^i(x)$ and of radius $\epsilon$ is an open covering of $K$. Since $K$ is compact, there must exist a finite subcovering. Therefore, we can choose $i_1 \ldots i_l$ such that $\{f^{i_k}(x)\}_{k=1}^{l}$ is $\epsilon$-dense. Now $i_1 \ldots i_l$ must be a subset of $[-n,n] \cap \mathbb{Z}$ for some $n$.