Solution Verification: If $\{I_k\}_{k = 1}^n$ (open, bounded intervals) cover $\mathbb{Q} \cap [0, 1]$, then $\sum_{k = 1}^n m^\ast(I_k) \geq 1$. I have a proof of Problem #8 from Royden & Fitzpatrick's Real Analysis but I am suspicious whether it works, because I can't find any similar approaches to the problem. So the set up is:

Let $\{I_k\}_{k = 1}^n$ be a finite collection of open, bounded, non-empty intervals (for $n \in \mathbb Z_+$) that covers $\mathbb Q \cap [0, 1]$. Then show that $\sum_{k = 1}^n m^\ast (I_k) \geq 1$.

I would appreciate if someone verified (i) my general approach to the problem and (ii) the proof itself. If the proof is too cumbersome, verifying my general approach is sufficient. Also, any improvements/simpler approaches are welcome.
Tentative Approach: My general idea is that $\{I_k\}_{k = 1}^n$ "essentially" covers the whole of $[0, 1]$. In fact, I believe: the set of uncovered points $[0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be a subset of the right (or left) endpoints of the $I_k$s. Is this correct or am I missing a simple counterexample? Assuming it is, we move on:
Tentative Proof Sketch: As the $I_k$s are open, bounded, non-empty intervals, write each as $(a_k, b_k)$ for reals $a_k, b_k$. So let $x \in [0, 1] \setminus \bigcup_{k = 1}^n I_k$ and note that $0 < x < 1$. Consider the two sets:
\begin{gather*}
 A_\ast(x) = \{a_k \in \mathbb R : x \leq a_k,\ 1 \leq k \in \mathbb Z \leq n\} \\
 B^\ast(x) = \{b_k \in \mathbb R : b_k \leq x,\ 1 \leq k \in \mathbb Z \leq n\}
\end{gather*} $A_\ast(x)$ is non-empty. Indeed, one of the intervals $I_j = (a_j, b_j)$ covers $1$ (for $1 \leq j \in \mathbb Z \leq n$). Hence, $x < 1 < b_j$; but then, $x \leq a_j$, otherwise $x \in I_j$ would be true. Similarly, $B^\ast(x)$ is also non-empty. So, as these are finite sets, $a_\ast := \min(A_\ast(x))$ and $b^\ast := \max(B^\ast(x))$ exist, and, furthermore $b^\ast \leq x \leq a_\ast$.
Now, in the case of $x = b^\ast$, we are done, as $x$ is then a right endpoint of one of the $I_k$s. The remaining case: $b^\ast < x$ is impossible. Indeed, for a contradiction, assume $b^\ast < x$. Then $\max\{b^\ast, 0\} < x \leq \min\{a^\ast, 1\}$, and, by the density of the rationals, there is some rational $q$ such that $\max\{b^\ast, 0\} < q < x \leq \min\{a^\ast, 1\}$. So, since $q \in \mathbb Q \cap [0, 1]$, some $I_k$ contains the $q$ i.e. $a_k < q < b_k$. In particular, we have $b^\ast < q < b_k$ so that $b_k \not\in B^\ast(x)$. Therefore $x < b_k$ by definition of $B^\ast(x)$, and $a_k < q < x < b_k$ as a result. But this means $x \in I_k$, which is not possible. Hence, every $x \in [0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be the right endpoint of some $I_k$ i.e. $x \in B := \{b_k \in \mathbb R : 1 \leq k \in \mathbb Z \leq n\}$. So $[0, 1] \setminus \bigcup_{k = 1}^n I_k$ must be a finite set $\{x_i\}_{i = 1}^{n'}$ (for $n' \in \mathbb N$), as $B$ is finite.
The rest of the proof is trivial and therefore omitted: I essentially throw in the singleton sets $[x_i, x_i]$ of uncovered points into our existing interval collection $\{I_k\}_{k = 1}^n$. The new collection obviously covers $[0, 1]$. So by properties of $m^\ast$ (finite subadditivity, monotonicity, interval-length preservation) the conclusion follows.
 A: Your argument is correct. Here is an alternative approach using induction.
Let us prove by induction the slightly more general result:

Let $\{I_k\}_{k = 1}^n$ be a finite collection of open, bounded, non-empty intervals (for $n \in \mathbb Z_+$) that covers $\mathbb Q \cap [a, b]$. Then show that $\sum_{k = 1}^n m^* (I_k) \geq b-a$.

Proof: Case $n=1$. Then $I_1 \supset \mathbb Q \cap [a, b]$ and it is clear that $\sum_{k = 1}^1 m^* (I_k)=m^* (I_1) \geq b-a$.
Now suppose we have proved the result for any collection of $n$ open bounded non-empty intervals.
Let $\{I_k\}_{k = 1}^{n+1}$ be a finite collection of open, bounded, non-empty intervals that covers $\mathbb Q \cap [a, b]$. Then there is at least one  $k\in \{1, \dots n+1\}$ such that $b \in I_k$. Since the collection is finite, we can re-index the collection so that this interval is numbered as $I_{n+1}$ (and such re-indexing does not change the value of $\sum_{k = 1}^{n+1} m^* (I_k)$).
Now, there are $c, d \in \mathbb R$, such that  $I_{n+1} =(c,d)$. Since $b \in I_{n+1}$, we have $c < b < d$.  We have now two possibilities:

*

*if $c< a$ then, $[a,b] \subset (c,d)= I_{n+1}$ and it is immediate that
$$\sum_{k = 1}^{n+1} m^* (I_k) \geq  m^* (I_{n+1}) >  b-a$$

*if $a \leq c$ then we have that $\{I_k\}_{k = 1}^{n}$ covers  $\mathbb Q \cap [a, c]$. Then, by the induction hypothesis, we have that $\sum_{k = 1}^n m^* (I_k) \geq c-a$. So we have
$$\sum_{k = 1}^{n+1} m^* (I_k) = \sum_{k = 1}^n m^* (I_k) + m^* (I_{n+1}) \geq (c-a)+(d-c) = d-a  > b-a$$
