Here is Prob. 19, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Let $E$ have finite outer measure. Show that if $E$ is not measurable, then there is an open set $O$ containing $E$ that has finite outer measure and for which $$ m^*(O \setminus E) > m^*(O) - m^*(E). $$
My Attempt:
As $E$ is not measurable, so by Theorem 11 (i) in Royden, there exists a real number $\epsilon_0 > 0$ such that, for every open set $O$ containing $E$, we have $$ m^* ( O \setminus E) \geq \epsilon_0. \tag{0} $$
Now as $m^*(E) < +\infty$ and as $\epsilon_0 > 0$, so $m^*(E) + \epsilon_0 > m^*(E)$, and thus by the definition of the outer measure there exists a countable collection $\left\{ I_{\epsilon_0, k} \right\}_{k=1}^\infty$ of non-empty bounded open intervals covering set $E$ for which $$ \sum_{k=1}^\infty l \left( I_{\epsilon_0, k} \right) < m^*(E) + \epsilon_0. \tag{1} $$ Let us put $$ O_{\epsilon_0} := \bigcup_{k=1}^\infty I_{\epsilon_0, k}. \tag{2} $$ Then by the choice of the $I_{\epsilon_0, k}$, we have $E \subset O_{\epsilon_0}$, which implies that $$ m^*(E) \leq m^* \left( O_{\epsilon_0} \right), \tag{3} $$ by the monotonicity of the outer measure, and also as $O_{\epsilon_0} = \bigcup_{k=1}^\infty I_{\epsilon_0, k}$, so by the subadditivity of outer measure we have $$ m^* \left( O_{\epsilon_0} \right) = m^* \left( \bigcup_{k=1}^\infty I_{\epsilon_0, k} \right) \leq \sum_{k=1}^\infty m^* \left( I_{\epsilon_0, k} \right) = \sum_{k = 1}^\infty l \left( I_{\epsilon_0, k} \right) < m^*(E) + \epsilon_0, $$ which implies that $$ m^* \left( O_{\epsilon_0} \right) < m^*(E) + \epsilon_0, $$ and as $m^*(E) < +\infty$, so we get $$ m^*\left( O_{\epsilon_0} \right) - m^*(E) < \epsilon_0. \tag{4} $$ Finally, as set $O_{\epsilon_0}$, being the union of open intervals, is an open set and moreover $O_{\epsilon_0} \supset E$, so from (0) and (4) above, we get $$ m^*\left( O_{\epsilon_0} \right) - m^*(E) < \epsilon_0 \leq m^* \left( O_{\epsilon_0} \setminus E \right), $$ which in turn implies $$ m^*\left( O_{\epsilon_0} \right) - m^*(E) < m^* \left( O_{\epsilon_0} \setminus E \right), $$ as required.
Is my proof correct and accurate in each and every detail? If so, then is the presentation also clear enough? Or, are there any issues?