Orbit of a homeomomorphism $g$ that is closed to the homeomorphism $f$

Question

Let $(X, d)$ be a metric space and $H(X)$ denotes the set of all homeomorphisms on $X$.

For $f, g\in H(X)$ define \begin{equation} d_0(f, g)= \sup \{d(f(x), g(x)), d(f^{-1}(x), g^{-1}(x)): x\in X\} \end{equation}

In my research $X=(\bigcup_{n\in\mathbb{Z}-\{0\}}X_n)\cup \{(-1,0), (1, 0)\}$ where $X_n=(1-\frac{1}{n}, [0, 2^{-n}])$ for $n\in\mathbb{N}$ and if $n\in \mathbb{Z}^{-}$, ($n< 0$), then $X_n=(-(1-\frac{1}{|n|}, [0, 2^{n}])$.

Consider a homeomorphism $f:X\to X$ such that $f(1, 0)=(1, 0)$, $f(-1, 0)= (-1, 0)$ and for $x\in X_n$, we have $f(x)\in X_{n+1}$. It is clear that $Fix(f)= \{(1, 0), (-1, 0)\}$.

Let $\delta>0$ be arbitrary small, $d_0(f, g)<\delta$ and $x\in X$.

What can say about orbit $g$ of $x$, $\mathcal{O}(x, g)=\{g^n(x): n\in \mathbb{Z}\}$?

Can I say that $g$ has atmost two fixed point?