I saw a few answers but I think that this proof is a slight variation
If $E_1$ and $E_2$ are measurable then $E_1\cup E_2$ is measurable.
Proof: it is sufficient to show that for all $A$, $m^*(A)\geq m^*(A\cap[E_1\cup E_2])+(A\cap[E_1\cup E_2]^C)$
$$A\cap(E_1\cup E_2)=(A\cap E_1)\cup (A\cap E_2)\subset (A\cap E_1)\cup (A\cap E_1^C \cap E_2)$$
Also $$A\cap(E_1 \cup E_2)^C=A\cap E_1^C\cap E_2^C$$
Therefore:
$$m^*(A\cap[E_1\cup E_2])+(A\cap[E_1\cup E_2]^C)\leq m^*(A \cap E_1)+m^*(A\cap E_1^C\cap E_2)+m^*(A \cap E_1^C \cap E_2^C)=_{(1)} m^*(A \cap E_1)+ m^*(A \cap E_1^C)=_{(2)}m^*(A)$$
a. why do $$(A\cap E_1)\cup (A\cap E_2)\subset (A\cap E_1)\cup (A\cap E_1^C \cap E_2) ?$$
b. Why there are equality $(1),(2)$ if $A=B\cup C$ then $m^*(A)\leq m^*(B)+m^*(C)?$