The first part of proving the collection of $\mu^*$-measurable sets is a $\sigma$-algebra is showing it is closed under finite (yes, finite) union. That is, for a set $\Bbb{X}$ and an outer-measure $\mu^*$ on $\Bbb{X}$, if $A,B$ are $\mu^*$-measurable sets, then $A \cup B$ is $\mu^*$-measurable as well. The proof starts like this:
$\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c) = \mu^*(E \cap A \cap B) + \mu^*(E \cap A \cap B^c) + \mu^*(E \cap A^c \cap B) + \mu^*(E \cap A^c \cap B^c)$.
Why is the second equality true? It actually means that a subset of a $\mu^*$-measurable set (here namely $E \cap A$ or $E \cap A^c$) is $\mu^*$-measurable, right?
It's using the $\mu^{\ast}$-measurability of $B$,
$$\mu^{\ast}(F) = \mu^{\ast}(F\cap B) + \mu^{\ast}(F\cap B^c)\,,$$
for the two sets $E\cap A$ and $E \cap A^c$.