We know a Markov Chain $\{X_n\}$ with a stationary distribution $\mu$ as its initial distribution is a stationary process, because if $X_0\sim \mu$ a stationary distribution, then we know for each $n$, $X_n \sim \mu$, then by markov property we know $(X_0, X_1, \cdots, X_n)\sim (X_m, X_{m+1}, \cdots, X_{m+n})$.
Then Durrett said a special case to keep in mind for counterexamples is the Markov chain $X_n: \Omega \rightarrow S=\{0,1\}$ with transition probability $p(0,1) = p(1,0) = 1$, and stationary distribution $\mu(0) = \mu(1) = {1}/{2}$. $(X_0, X_1, \cdots) = (0,1,0, \cdots) \text{ or } (1,0,1, \cdots)$ with probability $1/2$.
I am really confused, if the general statement is true, what is this counterexample implying, and why is this a counterexample?
Durrett is not saying that this Markov Chain is a counterexample to the definition of stationarity: in fact he is demonstrating that this chain satisfies stationarity.
What he is saying is that it is useful to keep this Markov chain in mind when picturing what stationarity means. In particular this is a commonly used counter example to distinguish between stationary distributions, and limiting distributions.
Recall that a Markov chain has a limiting distribution if
$$\pi_j = \lim_{n \rightarrow \infty} P_{ij}^n \qquad \forall \, i \in S,$$ exists. In particular that is if the limit does not depend on the starting state (and hence distribution) of the chain. Any chain that has a limiting distribution neccessarily is stationary (since $\pi$ can be shown to satisfy the stationarity property).
The converse however is not true: and this is what the counter example shows, since the limit above only exists if the chain is started from $\mu$, and not from an arbitrary distribution.
In general for finite, irreducible Markov chains