The author of this question might be more pleased with the following answer. Let $U$ be a uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + {\mathbf 1}(t=U)$, where ${\mathbf 1}$ denotes indicator function, is discontinuous at time $U$. However, for any choice of (fixed) times $t_i$, $i=1,...,n$, we have, almost surely, $W'(t_i) = W(t_i)$ for all $i$, and hence, trivially, $W'$ has the same distributional properties stated for $W$. Furthermore, if we define $W'$ by $W'(t) = W(t) + {\mathbf 1}(t \in UA)$, where $A$ is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$), then $W'$ is nowhere continuous (since $UA$ is dense in $(0,\infty)$); nevertheless, as before, almost surely $W'(t_i) = W(t_i)$ for all $i=1,...,n$ (since ${\rm P}(t \in UA) = {\rm P}(t/U \in A) = 0$).

Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2) The answer can be adapted to Lévy processes in general ($W$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions.

Finally, the author of this question ``wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text