= Solution
If $\sum_nm^*(A_n)=\infty$, the required inequality is automatic. Otherwise each $A_n$ has finite <Lebesgue outer measure>. Fix $\varepsilon>0$. By the defining <infimum>, for each $n\geq1$ choose a countable <interval> cover $\{J_{n,k}:k\geq1\}$ with
$$
\sum_{k\geq1}|J_{n,k}|<m^*(A_n)+\varepsilon2^{-n}.
$$
The family indexed by $(n,k)\in\mathbb N^2$ is countable and covers $\bigcup_nA_n$. Since all lengths are nonnegative, its total length is at most $\sum_nm^*(A_n)+\varepsilon$. Taking the infimum over covers of the union and then letting $\varepsilon\downarrow0$ gives
$$
\boxed{m^*\!\left(\bigcup_{n\geq1}A_n\right)\leq\sum_{n\geq1}m^*(A_n).}
$$
This proves <countable subadditivity of Lebesgue outer measure> directly from interval covers, without assuming that any $A_n$ is measurable.
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