In this MO post, the answers point out that the statement, "The reals are a countable union of countable sets" is consistent with $\textsf{ZF}$ (without choice). This question points out that in such models of $\textsf{ZF}$, Lebesgue measure $m$ as typically defined from its outer measure would yield $m(\mathbb{R}) = 0$, which of course is very bad. The issue, according to the answers in that question, is because the traditional proof of subadditivity for the outer measure requires Countable Choice $\textsf{AC}_\omega$.
However, I've heard that the statement "the countable union of countable sets is countable" ($\textsf{CUT}$) is weaker than $\textsf{AC}_\omega$. Is there, perhaps, a way to prove subadditivity of the Lebesgue outer measure $m^*: \mathscr{P}(\mathbb{R}) \to [0,\infty]$ given by $$ m^*(E) = \inf\left\{\sum_{n=1}^\infty b_n - a_n : E \subseteq \bigcup_{n=1}^\infty (a_n,b_n)\right\} $$ over $\textsf{ZF + CUT}$ (i.e., no $\textsf{AC}_\omega$)? Or even better, can we recover subadditivity in models of $\textsf{ZF}$ where $\mathbb{R}$ isn't the countable union of countable? If not, is full $\textsf{AC}_\omega$ required to get the subadditivity?
By the way, I am aware of an alternative to Carathéodory's approach which uses Borel-coded measures (described in Fremlin's Measure Theory, Volume 5), choice free. I am simply wondering if we can ever recover the approach in $\textsf{ZF}$ alone/plus extremely weak choice principles.
