Independence of the $X_i$ follows directly from independence of the $U_i$, since each $X_i$ is a measurable function of $U_i$ alone. For the marginal distribution, observe that the intervals $[(j-1)/n, j/n)$ are disjoint and each has length $1/n$, so $\mathbb{P}(X_i = x_j) = \mathbb{P}(U_i \in [(j-1)/n, j/n)) = 1/n$ for every $j$.