Assuming you don't mean continuous sample paths, the answer is no. For example, $X_t = $ the indicator of the rationals with probability 1, or on a space that is an atom of mass 1 if you prefer. Then there is only 1 set of positive probability and it doesn't work.
Since you do mean continuous sample path, the answer is 'yes', take $Y = \lbrace max |X_t| < A \rbrace $ which can be made to be of positive probability by the continuity of $X_t$. Then, by dominated convergence (convergence because of continuity, and dominated by the bound A) $f$ is continuous.