How to prove that $\text{null} \;T_1 \subset \text{null} \;T_2$ implies $\text{dim(range}\; T_1) \geq \text{dim(range}\ T_2)$? 
If $W$ is finite dimensional, $T_1, T_2 \in L(V,W)$ and $\text{null}\; T_1 \subset \text{null}\; T_2$ then $$\text{dim}( \text{range}\; T_1) \geq \text{dim}(\text{range}\; T_2)$$

I'm solving a problem in Linear Algebra Done Right and I'm still struggling with a problem. And in my approach I current facing this problem. I can prove it easily when $V$ is finite dimensional but things get ugly when I attempt to prove for more general case. Any help will be appreciated.

Let $w_1,\dots,w_r$ be a basis for $\operatorname{range}T_2$. Choose $v_1,v_2,\dots,v_r$ so that $T_2(v_i)=w_i$ for $i=1,2,\dots,r$. Finally define $u_1,\dots,u_r\in W$ by $u_i=T_1(v_i)$.
Since $(w_i)_{i=1}^r$ is a basis for $\operatorname{range}T_2$, any nontrivial linear combination of $(v_i)_{i=1}^r$ will not be in $\operatorname{null} T_2$, so it will not be in $\operatorname{null} T_1$, proving $(u_i)_{i=1}^r$ is linearly independent.  
