In this answer of mine, I used the obviously false statement that the Lebesgue measure is additive, while it is indeed only countably additive. But one could fix the argument if the following statement were true:
Let $\{A_i\}_{i\in I}$ be pairwise disjoint measurable subsets of $\Bbb R$, where $I$ is any set. Then $\sum_{i\in I} \lambda(A_i) \leqslant \lambda (\cup_{i\in I} A_i)$.
However, I do not even know if that statement is true, no matter how reasonable it looks like. Is there any obvious counter-example?
Obviously, if the subsets $\{A_i\}$ are assumed open, which is the case in the linked answer, the equality holds, since any open subset of the real line has at most countably many connected components. Nonetheless, this latter fact is in some sense what was asked to prove, with the constraint that one should not refer to the density and the countability of rationals. That is why here the statement refers to general subsets and not open subsets.
I managed to solve my problem thanks to the time I had to spend commuting in public transports. Let $J \subset I$ be a finite subset. By finite additivity, and since $\cup_{j\in J} A_j \subset \cup_{i\in I} A_i$, one has $$ \sum_{j \in J} \lambda(A_j) = \lambda\left( \cup_{j\in J} A_j\right) \leqslant \lambda\left( \cup_{i\in I} A_i\right). $$ It follows that $$ \sum_{i\in I} \lambda(A_i) = \sup_{J \subset I, |J| < +\infty} \sum_{j\in J} \lambda(A_j) \leqslant \lambda \left( \cup_{i\in I} A_i\right), $$ and the result I stated is true.