Here is Prob. 8, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Let $B$ be the set of rational numbers in the interval $[0, 1]$, and let $\left\{ I_k \right\}_{k=1}^n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^n m^*\left( I_k \right) \geq 1$.
My Attempt:
If one of the open intervals $I_k$ is unbounded, then by Proposition 1, Chap. 2, in Royden we have $$ \sum_{k = 1}^n m^*\left( I_k \right) = \sum_{k=1}^n l \left( I_k \right) = \infty, $$ where $l \left( I_k \right)$ denotes the length of the interval $I_k$ if $I_k$ is bounded and $+\infty$ if $I_k$ is unbounded, and thus the given inequality holds trivially. So let us assume that all these intervals are bounded.
If one of the intervals has length at least $1$, then again we have $$ \sum_{k = 1}^n m^*\left( I_k \right) = \sum_{k=1}^n l \left( I_k \right) \geq 1, $$ as required. So we assume that all these intervals are bounded with each having length less than $1$.
For each $k = 1, \ldots, n$, let us put $I_k := \left( a_k, b_k \right)$, where $a_k < b_k < a_k + 1$. We can assume without any loss of generality that $0 \in \left( a_1, b_1 \right)$ and $1 \in \left( a_n, b_n \right)$, that is, $a_1 < 0 < b_1$ and $a_n < 1 < b_n$.
Now $\sum_{k = 1}^n m^*\left( \left( a_k, b_k \right) \right) = \sum_{k = 1}^n l \left( \left( a_k, b_k \right) \right)$ is smaller when no two of these intervals overlap, and in that case we may assume that $$ a_1 < 0 < b_1 \leq a_2 < b_2 \leq \cdots \leq a_n < 1 < b_n. $$ However, if, for any $k= 1, \ldots, n-1$, we have $b_k < a_{k+1}$, then there will be infinitely many rational numbers in $[0, 1]$ that will not be covered by the collection $\left\{ I_k \right\}_{k=1}^n$. So we must have $$ a_1 < 0 < b_1 = a_2 < b_2 = \cdots = a_n < 1 < b_n, $$ and thus we obtain $$ \sum_{k=1}^n m^* \left( I_k \right) = \sum_{k = 1}^n \left( b_k - a_k \right) = b_n - a_1 > 1-0 = 1. $$ Thus the desired conclusion holds.
Is my proof satisfactory enough? If so, is my presentation clear enough too? Or, are there any loopholes?