The issue is that you have the incorrect probability space. Remember, your space $\Omega$ needs to be rich enough to deal with sample paths, not just sample points. For example, $(X_1,X_2,X_3) = (0,0,0)$, $(0,1,0)$, and $(0,1,1)$ should all have positive probability for the Markov chain you have defined, and yet your probability space only has two points! The correct probability space to consider is
$$\Omega :=\{0,1\}^{\mathbb Z}$$
with the product $\sigma$-field, and the unique probability measure $\mathbb P$ satisfying
$$\mathbb P\Big(\big\{\omega=\{\omega_i\}_{i\in\mathbb Z}\in\Omega:\omega_i = x_i\text{ for }i\in\{n,n+1,\ldots,n+m\}\big\}\Big) = \mu(x_n)\prod_{i=1}^mp(x_{i-1},x_i)$$
for each $n\in\mathbb Z$, $m\in\mathbb N$, and $(x_n,\ldots,x_{n+m})\in\{0,1\}^{m+1}$, where $\mu = \frac12\delta_{0} + \frac12\delta_{1}$ is the invariant distribution of your Markov chain. Your measure preserving transformation $\phi$ is just the shift operator, i.e.
$$\phi\Big(\{x_i\}_{i\in\mathbb Z}\Big) = \{y_i\}_{i\in\mathbb Z} \quad \text{where }y_i=x_{i+1}$$
and your observable $X$ is the just the value of the $0$ component of your sample point $\omega$, i.e.
$$X\Big(\{\omega_i\}_{i\in\mathbb Z}\Big) := \omega_0.$$
Under this notation, $X_n(\omega)=\omega_n$, so $\{X_n\}$ is a Markov chain with initial distribution $\mu$ and transition probabilities given by $p$. By construction, this is an ergodic Markov chain, and $\phi$ is an ergodic transformation.
More generally, if you replace $\{0,1\}$ with a countable state space $E$ and let $p$ be a transition kernel on $E$ with invariant distribution $\mu$, then the measure preserving transformation $\phi$ will be ergodic if and only if the induced Markov chain is ergodic, so the (seemingly different) definitions are consistent.
I don't have much experience of measure theory but I think I could understand what it means.
I assumed $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$. I noticed two points:
・$\phi$ is not bijective because $0.10101010..$ and $0.01010101..$ are both sent to $(0,0)$. $(\mathbb{R}-\mathbb{Q})/\mathbb{Z}$ may work better (but I haven't checked enough).
・I think you need to verify that the map $\phi$ preserve the measure because it is required to show "measurably isomorphic":
I think it means for any $U \subset X', \mu_{\mathbb{T}}(T_4(U)) = \mu_{\mathbb{T}^2}((T_2 \times T_2 )(\phi(U)))$
(but I am not so familiar with the topic so I am not sure.)
Best Answer
You are missing a hypothesis.
In Rokhlin's Lemma, the hypothesis is that you are working with a standard measure space, which your counterexample is not.