Prove that for any set $A$, $m^*\left(A \cap \bigcup_{k=1}^\infty E_k \right) = \sum_{k=1}^\infty m^*(A \cap E_k).$ 
Let $\{E_k\}_{k=1}^\infty$ be countable disjoint  collection of measurable sets. Prove that for any set $A$, $$m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right) = \sum_{k=1}^\infty m^*(A \cap E_k).$$

Suppose that $m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right)< \infty$. Now from subadttivity $$m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right) = m^*\left( \bigcup_{k=1}^\infty A \cap E_k  \right) \le \sum_{k=1}^\infty m^*(A \cap E_k).$$
Since $m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right)< \infty$ we have that $A \cap \bigcup_{k=1}^\infty E_k $ is bounded. Thus there exists a $G_\delta$ set $G$ such that $A \cap \bigcup_{k=1}^\infty E_k \subseteq G$ and $m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right) = m^*(G)$.
Now $$ \begin{align} m^*\left(A \cap \bigcup_{k=1}^\infty E_k  \right) &= m^*(G) \\ &\ge m^*\left(G \cap \bigcup_{k=1}^\infty E_k  \right) \\ &= m^*\left(\bigcup_{k=1}^\infty G \cap E_k  \right)\\  &= \sum_{k=1}^\infty m^*(G \cap E_k) \\ &\ge \sum_{k=1}^\infty m^*(A \cap E_k). \end{align}$$
What I'm worried about is that how come $ m^*\left(\bigcup_{k=1}^\infty G \cap E_k  \right)  = \sum_{k=1}^\infty m^*(G \cap E_k)$? This is exactly what we want to prove and it seems that we're using it out of the blue in the proof?

Notice the difference between outer measure $m^{*}$ and Lebesgue measure $m$.
We did not assume that $A$ is Lebesgue measurable. However, a $G_{\delta}$ is necessarily a Lebesgue measurable set, simply because it is a Borel set. A nice property of Lebesgue measure says that

If $\{A_{i}\}$ is a disjoint countable collection of Lebesgue
measurable sets, then $$m\left(\bigcup_{i=1}^{\infty}
 A_{i}\right)=\sum_{i=1}^{\infty}m\left(A_{i}\right).$$

But outer measure does not have such property.
