Prove that $|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff (a,b) \cap (c,d) = \emptyset$ This is from Section 2A, exercise 7 in Axler's MIRA book.

Suppose $a,b,c,d$ are real numbers with $a<b$ and $c<d$. Prove that
$$|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff (a,b) \cap (c,d) = \emptyset$$

I am having trouble coming with the right argument. Visually it makes a lot of sense that the intersection is empty, but I am having trouble using this intuition right now. For example, if I assume the intersection isn't empty, then how does that change the equality? In particular, how would the right side $(b-a) + (d-c)$ be affected as a result?
For reference, $||$ denotes the outer measure, so $|(b-a)| = b-a$ says that the outer measure of the open interval (a,b) is equal to b-a.

Updated with proof: Let $A_1 = (a,b)$ and $A_2 = (c,d)$. Take the union $A = A_1 \cup A_2$, where $A_1$ and $A_2$ are not necessarily disjoint. Then from elementary set theory, we can express $A$ as the union of disjoint sets $A = A_1 \cup A_2 \bigcup A_1 \cap A_2$. Then we have $$|A_1 \cup A_2 \bigcup A_1 \cap A_2 | = |A_1| + |A_2| + |A_1 \cap A_2|.$$ Then the original equality assumption implies that $A_1 \cap A_2 = \emptyset$ as desired.
 A: We have
$$
\begin{align}
& \, (b-a) + (d-c) - |(a,b) \cup (c,d)| \\
 =& \, |(a, b)| + |(c, d)| - |(a,b) \cup (c,d)|  \\
 =& \, |(a,b) \cap (c,d)|
\end{align}
$$
where the last identity is an  application of $|A \cup B| + |A \cap B| = |A| + |B|$.
It follows that
$$
\begin{align}
&|(a,b) \cup (c,d)| = (b-a) + (d-c) \\
\iff & |(a,b) \cap (c,d)| = 0 \\
\iff & (a,b) \cap (c,d) = \emptyset.
\end{align}
$$
The last equivalence holds because an open set has Lebesgue measure zero if and only if it is the empty set.
A: If $(a,b)\cap (c,d) \neq \emptyset$ then, by considering cases one can see that the outer measure of $(a,b)\cup (c,d)$ is not the same as $b-a + d-c$.
For the other side, assume they are disjoint, moreover assume that $b\leq c$, then by countable subadditivity, $\lvert (a,b)\cup (c,d)\rvert\leq b-a+d-c$. On the other hand $(a,d)=(a,b)\cup [b,c] \cup (c,d)$, so again by countable subadditivity, $d-a\leq \lvert (a,b) \cup (c,d)\rvert + (c-b)$. Taking $c-b$ to the other side we obtain the result.
