Proof involving rank of partitioned matrix. 
Prove that $$\operatorname{rank}\begin{bmatrix}     A & B  \\     0 & C  \end{bmatrix}\geq \operatorname{rank}A+\operatorname{rank}C.$$ 
Deduce that the rank of an upper triangular matrix is not less than
  the number of non-zero diagonal elements.


If I take the column bases of $A$ and $C$, the corresponding columns of the bigger matrix are Linearly Independent. 
This above idea is fairly intuitive and very helpful, but I am unable to write it down as a formal proof. Could someone help me express it in a clean and presentable form? 

For the deduction part, I see that any Upper Triangular Matrix can be expressed as  $     \begin{bmatrix}     A & B  \\     0 & C   \\     \end{bmatrix}$. 
Now, if I could show that: 
rank of $A \geq$ number of non-zero diagonal elements in A $\ \ \  \ \ \ \ \ -(1)$ 
rank of $C \geq$ number of non-zero diagonal elements in C $\ \ \  \ \ \ \ \ -(2)$ 
then the deduction logically follows.
The thing I find strange is that $A$ and $C$ are both themselves Upper Triangular Matrices. So effectively, both $(1)$ and $(2)$ are exactly what I started out to deduce in the first place. This basically puts me in a loop. I seem to be missing something and would like some help here. Thanks.

Hint. Let $r,s$ be the ranks of $A$ and $C$ respectively, $\{Ax_1,\ldots,Ax_r\}$ be a basis of the column space of $A$ and $\{Cy_1,\ldots,Cy_s\}$ be a basis of the column space of $C$. Now show that the vectors
$$
\pmatrix{Ax_1\\ 0},\ldots,\pmatrix{Ax_r\\ 0},\pmatrix{By_1\\ Cy_1},\ldots,\pmatrix{By_s\\ Cy_s}
$$
are linearly independent and they all reside in the image of $\pmatrix{A&B\\ 0&C}$.
