If $m$ is a regular Borel outer measure is it true that $m$ is topologically additive? If so what is a proof or a counterexample?

Definitions:

Topologically Additive: $X$ is a topological space, $m$ is a outer measure. $m$ is topologically additive iff $S,T \subseteq X$ separated by neighborhoods implies $m(S \cup T) = m(S)+ m(T)$.

Borel: $X$ topological space, $m$ outer measure on $X$. $m$ is Borel iff every open set is measurable

Regular: $X$ topological space, $m$ outer measure on X. $m$ is regular iff :

$K$ compact implies $m(K) < \infty$

$m(S) = \inf \{m(U) \colon S \subseteq U, U \text{ is open} \}$

For $U$ open, $m(U) = \sup \{m(K): K \subseteq U, K \text{ is compact}\}$