Prob. 10, Chap. 2, in Royden's REAL ANALYSIS: If, for some $\alpha>0$, $|a-b|\geq\alpha$ for all $a\in A$ and $b\in B$, then ... Here is Prob. 10, Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Let $A$ and $B$ be bounded sets for which there is an $\alpha > 0$ such that $\lvert a-b \rvert \geq \alpha$ for all $a \in A$ and $b \in B$. Prove that $m^*(A \cup B) = m^*(A) + m^*(B)$.

Of course, we have
$$
m^*(A \cup B) \leq m^*(A) + m^*(B), \tag{1}
$$
by virtue of Proposition 3, Chap. 2, in Royden.
How to prove the equality? In particular, how to prove the inequality $m^*(A) + m^*(B) \leq m^*(A \cup B)$?
For each $a \in A$, we have $B \cap (a-\alpha, a+\alpha ) = \emptyset$, and for each $b \in B$, we have
$A \cap (b-\alpha, b+\alpha) = \emptyset$. Thus of course we also have $A \cap B = \emptyset$.
How to make use of these facts in arriving at our desired equality?
PS:
Based on the comments below, here is my attempt at proving the desired inequality:

If $m^*(A \cup B) = \infty$, then we are done. Su let us assume that $m^*(A \cup B) < \infty$.


If either of $A$ and $B$ is countable, say set $A$, then we have $m^*(A) = 0$, and thus we have
$$
m^*(A) + m^*(B) = m^*(B) \leq m^*(A \cup B),
$$
as required. So let us assume that both the sets $A$ and $B$ are uncountable.


Then by definition $m^*(A \cup B)$ is the infimum of the set of all the sums $\sum_{k=1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty, bounded open intervals covering $A \cup B$.


Let $\left\{ I_k \right\}_{k=1}^\infty$ be any countable collection of non-empty, bounded open intervals covering $A \cup B$. Suppose  $I_k := \left( a_k, b_k \right)$ is any interval in this covering. Then we can sub-divide $I_k$ into $n_k$ disjoint sub-intervals $I_{k, i} := \left( a_{k, i}, b_{k, i} \right)$ such that, for each $i = 1, \ldots, n_k$, we have
(1) $b_{k, i} - a_{k, i} < \alpha$,
(2) no endpoints $a_{k, i}$ and $b_{k, i}$ are in $A \cup B$,  and
(3) $$
a_k = a_{k, 1} < b_{k, 1} = a_{k, 2} < b_{k, 2} = \cdots = a_{k, n_k} < b_{k, n_k} = b_k.
$$
Then of course we have the equality
\begin{align}
\sum_{i = 1}^{n_k} l \left( I_{k, i} \right) &= \sum_{i=1}^{n_k} \left( b_{k, i} - a_{k, i} \right) \\ 
&= b_{k, n} - a_{k, 1} \\ 
&= b_k - a_k \\ 
&=  l \left( I_k \right),
\end{align}
and thus
$$
\sum_{k=1}^\infty l \left( I_k \right) = \sum_{k=1}^\infty \sum_{i = 1}^{n_k} l \left( I_{k, i} \right). \tag{2} 
$$


Note that no open interval of length less than $\alpha$ can contain a point of set $A$ and a point of set $B$ at the same time, because each point of $A$ is at a distance of at least $\alpha$ from each point of $B$.


Thus we can partition the collection $\left\{ I_k \right\}_{k=1}^\infty = \left\{ \bigcup_{i=1}^{n_k}  I_{k, i} \right\}_{k=1}^\infty$ into two disjoint countable sub-collections  $\left\{ I_k^\prime \right\}_{k=1}^\infty$ and $\left\{ I_k^{\prime\prime} \right\}_{k=1}^\infty$ covering $A$ and $B$, respectively.


We by virtue of (2) above therefore obtain
$$
\begin{align}
m^*(A) + m^*(B) &\leq \sum_{k=1}^\infty l \left( I_k^\prime \right) + \sum_{k=1}^\infty l \left( I_k^{\prime\prime} \right) \\
&= \sum_{k = 1}^\infty l \left( I_k \right).
\end{align}
$$


But $\left\{ I_k \right\}_{k=1}^\infty$ was any covering of $A \cup B$ by a countable collection of non-empty, bounded open intervals. Thus we have shown that the number $m^*(A) + m^*(B)$ is a lower bound for the set of all the sums $\sum_{k=1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty, bounded open intervals covering $A \cup B$.  Thus it follows that
$$
m^*(A) + m^*(B) \leq m^*(A \cup B), 
$$
as required.

Is my proof correct in each and every detail? If so, is my presentation clear enough? Or, are there any issues?
Last but not least, is the assumption of boundedness for the sets $A$ and $B$  essential for our conclusion to hold?
 A: Actually, $m^*()$ in this chapter is not only an outer measure but also a Lebesgue outer measure. And more generally, if this is a metric outer measure, it satisfies the property of totally additive of disjoint sets.
By the definition in this book,
$m^*(A)$
is
$$m^*(A)=\inf \left\{\sum_{k=1}^∞ \mathscr l(I_k) \, | \, \mathsf A \subseteq \bigcup_{k=1}^∞ I_k \right\}$$
And due to the definition of the $\inf$, for any arbitrary $\epsilon > 0$, we could find a covering of $A \cup B$, say $\cup_{k=1}^∞ I_k^{A \cup B} $,  which satisfies
$$
\sum_{k=1}^∞ \mathscr l(I_k^{A \cup B})\le m^*(A \cup B) + \epsilon.
$$
And due to this is Lebesgue measure, we could let $\mathscr l(I_k^{A \cup B})\lt\alpha/2$ for any $k$. So none of $I_k^{A \cup B}$ could intersect with $A$ and $B$ at the same time.
Denote the covering of A, $\cup_{\{ I_k^{A \cup B} \cap A \neq \emptyset\}} I_k^{A \cup B}$ as $C_A$, and the covering of B, $\cup_{\{ I_k^{A \cup B} \cap B \neq \emptyset\}} I_k^{A \cup B}$ as $C_B$ .
Because of $\alpha \neq 0$ ,$C_A$ and $C_B$ are disjoint, then we have $$\sum_{k=1}^∞ \mathscr l(A\cap I_k^{A \cup B})  +\sum_{k=1}^∞ \mathscr l(B\cap I_k^{A \cup B}) \le  m^*(A \cup B) + \epsilon.$$
By the definition of $m^*(A)$ and $m^*(B)$, we have
$$
m^*(A) + m^*(B) \le \sum_{k=1}^∞ \mathscr l(A\cap I_k^{A \cup B})  +\sum_{k=1}^∞ \mathscr l(B\cap I_k^{A \cup B}) \le  m^*(A \cup B) + \epsilon 
$$
We could let $\epsilon$ limit to $0$, then we have the other direction of the equation.
A: My proof is based on the following very useful fact about outer Lebesgue measure:

Fix any $\delta >0$. In your definition of $m^*$ you can, without loss of generality, assume that length of the intervals that you use do not exceed $\delta$.

This is true because you can replace any longer interval with finitely many subintervals with length $\leq \delta$.
Fix a near optimal covering of $A \cup B$ by intervals of length at most $\delta$. Throw out any stupid interval that has zero intersection with $A \cup B$. The rest can be sorted into those touching $A$ and those touching $B$ -- without ambiguity because by choice of $\delta$ no interval of length $\delta$ or less can touch both $A$ and $B$.
Now you have coverings for A and B out of the covering for $A \cup B$. These give
$$
m^*(A) + m^*(B) \leq \text{sum of all $\ell(I)$} \simeq m^*(A\cup B) \, .
$$
