An outer measure is countable-additive on the measurable sets

You can show the finite case with finite induction on $n$. It works because the disjoint union is monotone increasing in terms of containment as $n$ increases. For the infinite case observe that for all $n$ $$ E\cap\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^\infty E\cap A_k\supseteq\bigcup_{k=1}^nE\cap A_k $$ so that $$ \mu^*(E\cap\bigcup_{k=1}^\infty A_k)=\sum_{k=1}^\infty \mu^*(E\cap A_k)\geq\sum_{k=1}^n\mu^*(E\cap A_k). $$ Since the LHS doesn't involve $n$, taking the limit finishes the proof!

More detail:

First I accidentally reversed the set names, see my edits above. Second, for the finite case, since each $A_k$ is measurable then for any set $E$ we have by induction: $$\begin{align*} \mu^*(E\cap\bigcup_{k=1}^nA_k)&=\mu^*((E\cap\bigcup_{k=1}^nA_k)\cap A_n)+\mu^*((E\cap\bigcup_{k=1}^nA_k)\cap A^C_n)\\ &=\mu^*(E\cap A_n)+\mu^*(E\cap\bigcup_{k=1}^{n-1}A_k)\\ &=\mu^*(E\cap A_n)+\sum_{k=1}^{n-1}\mu^*(E\cap A_k)\\ &=\sum_{k=1}^{n}\mu^*(E\cap A_k)\\ \end{align*} $$

If $m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) = \infty$, the reserve inequality holds trivially. If $m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) < \infty$, we can find a $G_\delta$ set $G$ that contains $A \cap \bigcup_{k=1}^\infty E_k$ and satisfies $$ m^*(G) = m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) $$ Since $G$ is measurable, we have \begin{align*} m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) &= m^*(G) \\ &\geq m^*\left(G \cap \bigcup_{k=1}^\infty E_k\right) \\ &= m^*\left(\bigcup_{k=1}^\infty (G \cap E_k)\right) \\ &= \sum_{k=1}^\infty m^*(G\cap E_k) \\ &\geq \sum_{k=1}^\infty m^*(A\cap E_k) \end{align*} where the fourth line follows because $G \cap E_k$ is measurable for all $k$ and the inequalities follow from the monotonicity of outer measure.