Consider a matrix
$$ X = \begin{bmatrix} A & B \\ 0 & D \end{bmatrix} $$
where the number of rows of each matrix is greater than that of columns. I know that $$\mbox{rank}(X) \ge \mbox{rank}(A) + \mbox{rank}(D)$$ holds. Then, can we say that $X$ has full column rank when both $A$ and $D$ have full column ranks?
Yes in the case where $A$ and $D$ are both square. Here we have $\det(X) = \det(A)\det(D)\neq 0$, so $X$ is non-singular. Thus $X$ has full-rank.
If $A$ and $D$ are rectangular, then this is not true. For instance, let $$X = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}.$$