I need some help with the proof as I am not sure if it is right.
Edit
After reading Botnakov N. comment, I gave the proof another shot.
Since the Markov chain is irreducible then there exists a unique stationary distribution. Assuming that the markov chain is reversible then the detailed balance equations hold:
$$\pi(i) p_{i j}=\pi(j) p_{j i}, \quad i, j \in S$$
Let D = [ $\pi(1)$, ...... , $\pi(n)$] and using the fact that P = AD, then:
$$ p_{ji} = {[AD]}_{ji} = a_{ji} * \pi(j)$$
I know that $a_{ji} = \frac{\pi(i)}{\pi(j)} * a_{ij}$ from the detailed balance equations. Hence:
$$ p_{ji} = a_{ji} * \pi(j) = \frac{\pi(i)}{\pi(j)} * a_{ij} * \pi(j) = \pi(i) * a_{ij} = p_{ij} $$
Thus, P is a symmetric matrix. But I am not sure how to conclude it.
Now, I assume that there exists a symmetric matrix P such that for all i,j in S, $p_{i j}= p_{j i}$ then the komologrov loop criterion is satisfied then the chain is reversible.
I am not sure how to show this in detailed
You write: "This implies that for all $n,\left(X_{0}, X_{1}, \ldots, X_{n}\right)$ has the same distribution as $\left(X_{n}, X_{n-1}, \ldots, X_{0}\right)$. therefore, $p_{i j}= p_{j i}$ and hence P is a symmetric matrix." No, $p_{ij} \ne p_{ji}$. It's not a corollary and moreover it's false. For example, consider a chain with transition matrix
\begin{pmatrix} \frac12 & \frac12 \\ \frac13 & \frac23 \end{pmatrix} Here $p_{12} \ne p_{21}$ and an easy computation gives that $\pi_1 = \frac25, \pi_2 = \frac35$ for probabilities of stationary distribution. Hence $\pi_1 p_{12} = \pi_2 p_{21}$ and the chain is reversible, q.e.d.
Remark In case when there are only $2$ states every chain with stationary distribution is reversible, because $\pi_1 p_{11} + \pi_2 p_{21} = \pi_1$ (as $\vec{\pi}$ is stationary) and it follows that $$\pi_1 p_{12} = \pi_1 (1-p_{11}) = \pi_1 - \pi_1 p_{11} = (\pi_1 p_{11} + \pi_2 p_{21}) - \pi_1 p_{11} = \pi_2 p_{22}.$$