Here $m^*(A)$ denotes the outer measure of $A$.
I know the proposition is true for measurable $A$, see
If $X \subset \mathbb R$ is measurable then for every $\alpha \in (0, \mu(X))$ there exists $X_\alpha \subset X$ such that $\mu(X_\alpha) = \alpha$
But is it true if $A$ is not measurable? The nuance here is that we can't deduce \begin{align} \lim_{x\to +\infty}m^*(A\cap[-x,x])=m^*(A), \end{align} as easily as when $A$ is measurable.
Edit:
The problem is solved.
Since continuity from below also holds for outer measure:
\begin{align}
A_n \nearrow A\implies\mu^*(A_n) \nearrow \mu^*(A),
\end{align}
we can see that the proof is essentially the same as that for measurable sets.
See
Continuity of outer measure induced by measure from below
Note that $\mu^*(A) \le \sum_{n\in \mathbb{Z}}\mu^*(A\cap [n, n+1))$. However, since the sets $[n,n+1)$ are measurable we have $\sum_{n=a}^{b-1} \mu^*(A\cap [n,n+1)) = \mu^*(A\cap [a,b))$. Therefore $$\lim_{n\to \infty} \mu^*(A\cap [-n, n])= \mu^*(A)$$
Note: More generally, if $A_n \nearrow A$ then $\mu^*(A_n) \nearrow \mu^*(A)$, the proof is somewhere on the site, with a beautiful idea.
$\bf{Added:}$ The proof of the last equality, from an answer on this site, can't locate it now.
Take $D_n$ measurable, $D_n \supset A_n$, and $\mu^*(A_n) = \mu(D_n)$ (a measurable hull). Now we can make $D_n$ also increasing: instead of $D_n$, take $D_n\colon=\cap_{m\ge n} D_m$ ( a wonderful idea, not mine). Now we get $$\mu^{*}(A_n) = \mu(D_n) \nearrow \mu(\cup_{n} D_n) \ge \mu^{*}(A)$$