If you mean that if $\mu^*$ is a metric outer measure, then all Borel sets are $\mu^*$-measureable, the answer is yes:
Since $\mu^*$ is a metric outer measure, $\mu^∗(E ∪ F) = \mu^∗(E) + \mu^∗(F)$ whenever dist $(E, F) > 0.$
Let $U$ be open in the metric space $X$. Define $U_n=\left \{ x ∈ U : \text {dist} (x, X\setminus U) >1/n \right \}.$
Take $D_n=U_{n+1}\setminus U_n,\ $so that $U\setminus U_n=\bigcup^{\infty}_{i=n} D_i\ $ and dist $(D_i, D_j ) ≥1/(i + 1)−1/j> 0$ whenever $i + 2 ≤ j.$
Thus, since $\mu^*$ is a $metric$ measure, we have, for all $A\subseteq X \ ;\ $(wlog $\mu^*(A)< \infty),$
$\mu^*(A\cap D_1)+\cdots +\mu^*(A\cap D_{2n-1})= \mu^*(A\cap (D_1\cup\cdots\cup D_{2n-1})\le \mu^*(A),\ $and similarly for the even $1\leq i\le n.$
Therefore,
$\sum_{i=1}^{\infty}\mu^*(A\cap D_i)\le 2\mu^{*}(A)<\infty.$
So now we can say that
$\mu^*(A\cap (U\setminus U_n))=\mu^*(A\cap (\bigcup^{\infty}_{i=n} D_i))\le \sum_{i=n}^{\infty}\mu^*(A\cap D_i)\to 0\ $ as $n\to \infty.$
Using again, the fact that $\mu^*$ is a metric outer measure, we compute
$\mu^*(A\cap U_n)+\mu^*(A\setminus U)=\mu^*((A\cap U_n))\cup (A\setminus U))\le \mu^*(A),\ $and finally
$\mu^*(A\cap U)+\mu^*(A\setminus U)\le \mu^*(A\cap U_n)+\mu^*(A\cap (U\setminus U_n))+\mu^*(A\setminus U)\le \mu^*(A)+\mu^*(A\cap (U\setminus U_n))\ $
and it remains only to pass to the limit to see that the Carotheodory condition is satisfied.