$X(n)$ is a Markov chain on $X$, a separable, complete metric space, is the following chain Feller? i.e for every $f\in C(\mathcal X, \mathbb R)$, the map $x\mapsto \mathbb E ( f(X(1))| X(0)=x)$ is continuous? And, for every $X(0)=x\in \mathcal X$ and for any $f\in f\in C(\mathcal X, \mathbb R)$, do we have $\lim\limits_{k\to \infty}\frac{1}{k+1}\sum\limits_{j=0}^{k}f(X(j))=\int f(x)\pi(dx)$, forsome $\pi$ unique invariant measure.
We can think of the chain is generated by $X(n)=f_{\theta}(X(n-1))$ where $\theta\in \Theta$, $f_\theta$ is a family of Lipschitz maps on $X$.$\Theta$ is finite set say $\{1,\dots,p\}$, $m$ is a probability distribution on $\Theta$ such that probability of selecting $f_\theta$ is $m(\theta)$.
For $X$ compact I know, a result kind of related to this, ref1
The chain is Feller, since $$E[f(X(1)) \mid X(0)=x] = \sum_{\theta \in \Theta} m(\theta) f(f_\theta(x))$$ and the right side is just a finite linear combination of continuous functions of $x$.
It does not have to have be ergodic, since, for instance, such a chain can simply be uniform motion to the right, i.e. on the state space $\mathbb{R}$, $X(n) = n$ (take $\Theta = \{\theta_0\}$ to have one point, and $f_{\theta_0}(x) = x+1$). Then the limit you are taking is $\lim_{k \to \infty} \frac{1}{k+1} \sum_{j=0}^k f(j)$ and you can certainly find continuous functions $f$ for which this diverges (something like $f(x) = \sin(\log x))$ will probably work).