Let $(B_t)$ be a standard Brownian motion and consider
$X_t^x = \begin{cases} B_t + x &\mbox{ } x \neq 0 \\ 0 & \mbox{ } x = 0 \end{cases}$
Then one can show, that $(X_t, \mathcal{F}_t^B, P_x)$ is a Markov family and hence one can construct a Markov process. To show that this process is not strong Markov, let's consider $\tau := \{ t > 0 | X_t = 0 \}$. Then $X_{\tau} = 0$ almost surely. Let's assume that the strong Markov property holds, than it would follow for $x \neq 0$:
(1) $P_x(X_{\tau + h} = 0 | \mathcal{F}_{\tau}) = P_{X_{\tau}}(X_h = 0) = 0$,
which contradicts
(2) $P_0(X_{\tau + h} = 0 | \mathcal{F}_{\tau}) = 1$.
My problem now with this example is that I cannot really see, why equation (1) is a contradiction to equation (2).