I give an elementary solution to Problem 1 in $\mathbb{R}^n$ in my book *Measure Theory and Functional Analysis* (Proposition 2.16, p. 48). Here is the one-dimensional version.

I guess it's clear that the length function defined on finite disjoint unions of intervals is finitely additive. Thus for each $N \in \mathbb{N}$ we have
$$\sum_{n=1}^N l(I_n) = l\left(\bigcup_{n=1}^N I_n\right) \leq l(I),$$
and taking $N \to \infty$ yields $\sum l(I_n) \leq l(I)$.

For the reverse inequality, let $\epsilon > 0$. Let $K$ be a closed interval contained in $I$ whose length is at least $l(I) - \epsilon$ and for each $n$ let $J_n$ be an open interval that contains $I_n$ and whose length is at most $l(I_n) + \frac{\epsilon}{2^n}$. Then the $J_n$ cover $K$, so by compactness finitely many of them cover $K$. Thus $K$ is contained in $\bigcup_{n=1}^N J_n$ for some $n$. Defining $K_n = (K \cap J_n) \setminus (J_1 \cup \cdots \cup J_{n-1})$ then yields finitely many disjoint intervals $K_1$, $\ldots$, $K_N$ whose union is $K$. Thus
$$l(I) \leq l(K) + \epsilon = \sum_{n=1}^N l(K_n) + \epsilon \leq \sum_{n=1}^N l(J_n) + \epsilon \leq \sum_{n=1}^N l(I_n) + 2\epsilon.$$
Taking $\epsilon \to 0$ yields $l(I) \leq \sum l(I_n)$.