Let $(X, d)$ be a compact metric space . Given $\delta>0$ and a subset $A\subseteq X$ we define \begin{equation} A^\delta=\{x\in X: d(x, a)<\delta \text{ for some } a\in A\} \end{equation}
Define the Hausdroff metric between $A, B\subseteq X$ by
\begin{equation} d_H(A, B)=\inf \{ \delta: A\subseteq B^\delta, B\subseteq A^\delta\} \end{equation} $f:X\to X$ is a minimal homeomorphis, if every point of $X$ has a dense orbit.
In a paper author claim that if $f:X\to X$ is a minimal homeomorphis, then for every $\epsilon>0$, there is $\delta>0$ such that for every homeomorphim $g:X\to X$ with $d(f(x), g(x))<\delta$, for all $x\in X$, we have $d_H(X, \overline{O_g(z)})<\epsilon$ for all $z\in X$, where $O_g(z)=\{g^n(z)\}_{n\in\mathbb{Z}}$.
For $x\in X$, and $n\in\Bbb N$, let $$O_n(x)=\{\,f^k(x)\mid 1\le k\le n\,\} $$ and $$U_n(x)=\{\,y\in X\mid d(y,O_n)<\tfrac12\epsilon\,\}. $$ By minimality of $f$, we have $\bigcup_n U_n(x)=X$ and then by compactness $X=U_n(x)$ for some $n=n(x)$. Then $d_H(X,O_{n(x)}(x))<\frac12\epsilon$, and as $d_H(X,O_{n(x)}(x'))$ is continuous in $x'$, we see that $n(x')\le n(x)$ for all $x'$ in an open neighbourhood $V_x$ of $x$. The $V_x$ cover $X$, so by compactness there is a finite subcover. If $N$ is the maximum over the finitely many $n(x)$ involved in this subcover then we find that $$d_H(X,O_N(x))<\frac12\epsilon $$ for all $x\in X$.
Let $\delta_0=\frac12\epsilon$. By continuity of $f$ and compactness once again, there recursively exists $\delta_{n+1}>0$ such that $d(x,x')<\delta_{n+1}$ implies $d(f(x),f(x'))<\frac12\delta_n$. Let $\delta=\frac12\min\{\delta_0,\ldots,\delta_N\}$. Now if $d(f,g)<\delta)$, then for any $z$, we can show by induction that $d(f^k(z),g^k(z))<\delta_{N-k}$ for $k=1,\ldots, N$. In particular, $d(f^N(z),g^N(z))<\frac12\epsilon$ and the claim follows.