Symmetric difference equivalence relation

This boils down to a set theoretic identity: $A \setminus C \subseteq (A \setminus B) \cup (B \setminus C)$. Indeed, let $x \in A \setminus C$, then $x \in A$ and $x \notin C$. If $x \in A \setminus B$, then we're done. If $x \notin A \setminus B$, then $x \in B$ (since the only way that $x \in A$ and $x \notin A \setminus B$ is that $x \in B$). Now we know that $x \in B$ and $x \notin C$, so $x \in B \setminus C$. Therefore $x \in (A \setminus B) \cup (B \setminus C)$. Similarly, you can prove that $C \setminus A \subseteq (C \setminus B) \cup (B \setminus A)$. You should now be able to convince yourself that $A \Delta C \subseteq (A \Delta B) \cup (B \Delta C)$ and hence use the properties of measures to finish off the inequality you're looking for.