Found this proof in Inder K. Rana - An Introduction to measure theory and integration
The author wants to prove that $\lambda(I = \cup_{i=1}^{\infty} I_n) = \sum_{i=1}^{\infty} \lambda(I_i)$ where $\lambda: \mathcal{I} \to [0, \infty]$ and $\mathcal{I}$ is the set of all possible intervals of the real line.
Now in the line before the last:
why there are inequalities? - it seemed to me it should be like this:
$\lambda(I = \cup_{i=1}^{k} I_n) = \sum_{i=1}^{k} (b_i - a_i) = b-a$
The emphasis for the inequality is to take advantage of the fact that limits preserve it.
That is to say given a sequence of numbers $(a_n)$ and a constant $C$, $$a_n\le C\implies \lim a_n\le C$$ provided the limit exists.
The line before last uses $C=\lambda(I)$ and $(a_k)=(\sum_{n=1}^k\lambda(I_n))$. Also note the limit exists by the monotone convergence theorem.