In particular, I need to prove the following: $d(A_1,B_1)+d(A_2,B_2)\geq d(A_1\cup A_2,B_1\cup B_2)$ and $d(A_1,B_1)+d(A_2,B_2)\geq d(A_1\cap A_2,B_1\cap B_2)$.
Reading from Baby Rudin, it seems these require knowing properties of symmetric differences, namely $S(A_1,B_1) \cup S(A_2,B_2) \text{ contains } S(A_1\cup A_2,B_1\cup B_2), S(A_1\cap A_2,B_1\cap B_2), \text{ and }S(A_1\setminus A_2,B_1\setminus B_2)$
By the definition of an outer measure(particularly monotonicity and subadditivity), it suffices to show that $$S(A_1 \cup A_2, B_1 \cup B_2) \subseteq S(A_1, B_1) \cup S(A_2, B_2).$$
If $x \in S(A_1 \cup A_2, B_1 \cup B_2)$, then WLOG we can assume $x \in (A_1 \cup A_2) \setminus (B_1 \cup B_2)$. Then $x \in A_1$ or $x \in A_2$ so again we can ssume WLOG that $x \in A_1$. Since $x \in A_1$ and $x \notin B_1 \cup B_2$ so $x \notin B_1$. Hence. $x \in A_1 \setminus B_1 \subseteq S(A_1,B_1). $
The other inequality is very similar. No assumption of measurability is necessary.