I understand the majority of this solution, it's just I don't understand why I have to use both $\epsilon_1 $ and $\epsilon_2 $ rather than just $\epsilon$. I understand that i'm working with $|x_1x_3|$ rather than $|x_1x_2|$ but because the statement holds $\forall \epsilon>0$ is this not just an unneccesary complication?
It is true that since the inequality holds for arbitrary $\epsilon$ we could take both $\epsilon_{1}$ and $\epsilon_{2}$ to be equal but I think the author was being careful in their estimates so as to have as much control over the inequality as possible. In general it could be that no choice of equal parameters gives the desired inequality.
Notice that if $\epsilon_{1}=\epsilon_{2}$ then notice that we need to choose $\epsilon$ so that
$4-8\epsilon>0$, $10-\frac{2}{\epsilon}>0$ and $35-\frac{6}{\epsilon}>0$.
This amounts to having $\epsilon<\frac{1}{2}$ and $\epsilon>\frac{1}{5}$ and $\epsilon>\frac{6}{35}$ which is satisfied by $\epsilon=\frac{1}{3}$.