This question was already asked here. Homogenous Markov chains won't work because the probability distributions of $X_t$ and $X_s$ are mutually absolutely continuous for every positive $t$ and $s$. For a homogenous diffusion with nonzero diffusion term, $P[X_{2t}\in A]\ne0$ would require that $P_y[X_t\in A]\ne0$ for $y$ in a set of positive measure with respect to the probability distribution of $X_t$ while $P_x[X_t\in A]=0$. Unless I am missing something, this is impossible.
The idea of the proof for finite Markov chains is as follows. Assume that $P_x[X_t\in A]$ is positive for a given positive $t$. The transitions of the chain can be wriiten as $$ P_y[X_t=z]=(\mathrm{e}^{tQ})(y,z), $$ where $Q$ is the infinitesimal generator of the process. Choose $u$ such that $u\ge -Q(y,y)$ for every state $y$, then $uI+Q$ has nonnegative coefficients. Now, $$ P_x[X_t\in A]=\mathrm{e}^{-ut}(\mathrm{e}^{t(uI+Q)})(x,A) $$ is a linear combination with positive coefficients of nonnegative terms $(uI+Q)^n(x,A)$, hence $(uI+Q)^n(x,y)>0$ (uI+Q)^n(x,y)$ is positive for at least one given $n$.
Now, for every positive time $s>0$, s$, $P_x[X_s\in A]$ is also a linear combination with nonnegative coefficients of $(uI+Q)^k(x,y)$, in particular $P_x[X_s\in A]\ge\mathrm{e}^{-us}(s^n/n!)(uI+Q)^n(x,A)$ hence A]$ is at least $P_x[X_s\in A]>0$\mathrm{e}^{-us}(s^n/n!)(uI+Q)^n(x,A)$, hence is positive.
