 Solution

For each nonnegative rational $q$, equality of the two versions gives $\mathbb{P}(X_q=Y_q)=1$. Intersect these countably many full-probability events with the full-probability event on which both paths are càdlàg. Call the resulting event $A$; then $\mathbb{P}(A)=1$.

Fix $\omega \in A$ and any real $t\ge 0$. Choose rational numbers $q_n>t$ decreasing to $t$. Right continuity givesThe same event $A$ works for every $t$, because the argument is pathwise after $\omega$ is fixed. The stochastic processes are therefore indistinguishable. This proves that càdlàg versions are indistinguishable; the left limits are not needed for this implication, since right continuity alone suffices.