A Markov process which is not a strong markov process?

A standard example is Exercise 6.17 in Michael Sharpe's book General theory of Markov processes. The process stays at zero for an exponential amount of time, then moves to the right at a uniform speed.

Consider the following continuous Markov process X, starting from position x

1. if x = 0 then X_t = 0 for all times.
2. if x ≠ 0 then X is a standard Brownian motion starting from x.

This is not strong Markov (look at times at which it hits zero).

Let $X(t) = f(W(t) + \pi)$, where $W(t)$ is a standard Wiener process and $$f(x) = \begin{cases} (x,0), & x\leq 0 \\\ \\\ (\sin x,1-\cos x), & 0 < x < 2\pi \\\ \\\ (x-2\pi,0), & x\geq 2\pi \end{cases} $$ is a map from $\mathbb R$ to $\mathbb R^2$. $X(t)$ is an $\mathbb R^2$-valued Markov process on $\mathbb R_+$ which is not strongly Markovian. See "A Modern Approach to Probability Theory" by Fristedt and Gray (1997, pp. 626–627).

If the time set is discrete, the ordinary Markov property implies the strong Markov property.