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.

Since you can prove it for $V$ finite-dimensional, the easiest way out is to look at the induced linear maps on the quotient $V/\ker T_1\to W$.
