I’m looking for a simple example of a wide-sense stationary but not strict-sense stationary stochastic process. So far:
Let $X$ be a stochastic process with discrete time. For $n \in \mathbb{N}_0$, we define $X_{2n}$as a fair random choice out of $\{-1, 0, 1\}$ and $X_{2n+1} := |X_{2n}|-\frac{2}{3}$.
With $$\mathbf{E}\bigl[X_{2n+1}\bigr] = \frac{1}{3}\cdot\frac{2}{3} - \frac{2}{3}\cdot\frac{1}{3} = 0 $$ and obviously $\mathbf{E}\bigl[X_{2n}\bigr] = 0$, the mean function $m_X(n) = \mathbf{E}[X_n] = 0$ and therefore constant.
Note that $X_{2n+1} = \frac{1}{3}$ for $X_{2n} \neq 0$. Therefore the random variable $X_{2n} X_{2n+1}$ is a fair random choice out of $\bigl\{-\frac{1}{3}, 0, \frac{1}{3}\bigr\}$ and has zero mean.
With the results for the means we see that $$\mathbf{Cov}\bigl[X_{2n}, X_{2n+1}] = \mathbf{E}\bigl[X_{2n}X_{2n+1}\bigr] - \mathbf{E}\bigl[X_{2n}\bigr]\mathbf{E}\bigl[X_{2n+1}\bigr] = 0 - 0= 0\,.$$ Since any other pairing is even independent, $X$ is a stochastic process consisting of uncorrelated random variables.
Therefore the covariance function $$K_X(n, m) = \mathbf{Cov}\bigl[X_n, X_m] = \begin{cases} 1 & \text{for } n = m \\ 0 & \text{for } n \neq m\end{cases}$$ depends only on the lag $|n - m|$, and $\mathbf{E}\bigl[X_n^ 2\bigr]<\infty$.
$\Rightarrow$ $X$ is wide-sense stationary.
$X$ is not strict-sense stationary. The joint probability distribution of $(X_{0}, X_{2})$ is not the same as the joint probability distribution of $(X_{1},X_{3})$: \begin{align}\mathbf{P}\bigl[(X_{0},X_{2}) = (1,1)\bigr] &= \frac{1}{9} \\ \mathbf{P}\bigl[(X_{1},X_{3}) = (1,1)\bigr] &= 0 \,.\end{align}
Question: Is the example and the derivation correct?
Yes, this derivation is correct. More generally, if $F$ and $G$ are distinct probability distributions with the same first two moments, independently picking the even numbered $X_n$ from $F$ and the odd numbered $X_n$ from $G$ will be an example of what you want.