I have derived the inequality $$m_*(E_1\times E_2)\leq (m_*(E_1)+\epsilon)(m_*(E_2)+\epsilon).$$ To complete the proof, I proceed to make $\epsilon$ tend to zero. But I have one question here. Is it necessary to consider the case that both $E_1$ and $E_2$ have positive exterior measure? Why not let $\epsilon\to 0$ once we obtain the inequality? Thank you.
On
If $m_*(E_1)=0$, and $m_*(E_2)=\infty$, then we must prove $m_*(E_1\times E_2)=0$. However, one cannot attain this aim by letting in $$m_*(E_1\times E_2)\leq [m_*(E_1)+\epsilon][m_*(E_2)+\epsilon],$$ for the product on the right may go to infinity as $\epsilon\to 0$. Hence, we need to be careful if one of this two sets has exterior measure $0$.
It seems that the outer measure the author is working with is finite on bounded sets. (Can you confirm, please?)
If that's correct and if you have a close look at the reasoning for the special case you may figure out why some additional argument is necessary. In the discussion of the special case that one set has zero outer measure, the author is using the result of the original reasoning without hesitation for bounded sets. This makes sense to me only if he can conlude that then the second factor has finite measure.
The problem is that one set might have infinite measure, and $0\times \infty$ is undefined (wheras $a\times \infty = \infty$ for every $a>0$).