Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$

Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$, where $A,B \in M_{m\times n}(\mathbb{F})$.

I'm trying to think in terms of linear transformations.
We can define $T_a, T_b:\mathbb{F}^n\rightarrow \mathbb{F}^m$.
I know that $\dim_{\mathbb F}\operatorname{Im} T_a, \dim_{\mathbb F}\operatorname{Im} T_b \le m$.

What should I do next?

Hint: It suffices to prove that $C(A+B)\subseteq C(A)+C(B)$, because $$\begin{align} C(A+B)\subseteq C(A)+C(B) &\implies \dim \left(C(A+B)\right)\leq \dim \left(C(A)+C(B)\right)\\ &\implies \text{rank}(A+B) \leq \dim(C(A))+\dim(C(B))-\dim(C(A)\cap C(B)).\end{align}$$

How about this: Let $rk A: =a$ , $rk B:=b$. Now we can do Gaussian elimination on both, to end with two matrices $A'$, $B'$ with , respectively, $a$ and $b$ non-zero rows. But the sum $A'+B'$ will have at most max{a,b} nonzero rows. Then $a+b \geq$ Max{$a,b$}