Upfront, I haven't studied mathematics in many months so I decided to pick up where I left off with aspirations of going back to school for a masters degree in mathematics. I am working with Royden's/Fitzpatrick's Real Analysis (4th Ed.) to which a free download can be found here (a safe/valid link as of now, as I just re-downloaded it to be sure the link isn't broken after scanning the link for anything harmful using Google Safe Browsing). Anyways, I came across a problem and I need help in regards to verification whether my proof is correct or not. The problem that follows (stated exactly as it is given) is problem 8 from chapter 2 within Royden's/Fitzpatrick's book (problem 8 from section 2 of chapter 2, on page 34, of the book to be precise).
Problem: Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\big\{I_{k}\big\}_{k=1}^{n}$ be a finite collection of open intervals that covers $B$. Prove that ${\displaystyle{\sum_{k=1}^{n}m^{*}(I_{k})}}\geq 1$.
Proof: Proceeding by induction on $n\in\mathbb{N}$, we let $n=1$. So $I_{1}$ is an arbitrarily fixed open interval that covers $B=\mathbb{Q}\cap[0,1]$. This means that $B\subsetneq[0,1]\subset I_{1}$, since $B\subset I_{1}$. That said, we have $m^{*}(I_{1})\geq m^{*}\big([0,1]\big)=\ell\big([0,1]\big)=1$ by monotonicity of $m^{*}$ and the base case is true.
$~~~~~~~$We now let $n\in\mathbb{N}\backslash\{1\}$, and assume that given $\big\{I_{k}\big\}_{k=1}^{n}$ as a finite collection of open intervals that covers $B$, then ${\displaystyle{\sum_{k=1}^{n}m^{*}(I_{k})}}\geq 1$ holds as the induction hypothesis. Let $\big\{I_{k}\big\}_{k=1}^{n+1}$ be an arbitrarily fixed collection of open intervals that covers $B$. First, define $I_{n+1}:=J_{n+1}$. Now, if $\big\{I_{k}\big\}_{k=1}^{n}$ is a cover for $B$, define $I_{k}:=J_{k}$ for each $k\in\big\{1,2,...,n\big\}$. Otherwise, if $\big\{I_{k}\big\}_{k=1}^{n}$ is not a cover for $B$, then let $\big\{E_{k}\big\}_{k=1}^{+\infty}$ be a countable collection of open intervals that covers $[0,1]$. By the Heine-Borel theorem, there exists a finite subcover $\big\{E_{k}\big\}_{k=1}^{m}$ of $[0,1]$ as well as $\big\{E_{k}\big\}_{k=1}^{m}$ is a finite collection of open intervals that covers $B$. If $m<n$, then define $E_{k}:=J_{k}$ for each $k\in\big\{1,2,...,m\big\}$ and define $J_{k}:=(0,1)$ for each $k\in\big\{m+1,m+2,...,n\big\}$. If $m=n$ define $E_{k}:=J_{k}$ for each $k\in\big\{1,2,...,m\big\}$. If $m>n$, define $E_{k}:=J_{k}$ for each $k\in\big\{1,2,...,n-1\big\}$ and define $J_{n}:=(-1,2)$. In any of the cases mentioned, we have that both $\big\{J_{k}\big\}_{k=1}^{n}$ and $\big\{J_{k}\big\}_{k=1}^{n+1}$ are each a finite collection of open intervals that cover $B$. This now implies that:$${\displaystyle{\sum_{k=1}^{n+1}m^{*}(J_{k})=\sum_{k=1}^{n}m^{*}(J_{k})+m^{*}(J_{n+1})\geq\sum_{k=1}^{n}m^{*}(J_{k})\geq 1}},$$
by the induction hypothesis and since $m^{*}(J_{n+1})\in[0,+\infty]$ (such that $J_{n+1}$ could be empty, bounded or unbounded) concluding the proof.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$
When I returned to this problem, I forgot how I originally proved it. Then I noticed how things started to fall into place when I thought to use induction. At first, I thought we could (really easily) fix the collection of intervals covering $B$ from the inductive hypothesis and just throw any open interval $I_{n+1}$ (such that $\big\{I_{k}\big\}_{k=1}^{n}$ would still cover $B$) into the mix. Then I realized that might not be right since we don't know if $\big\{I_{k}\big\}_{k=1}^{n}$ is still a cover of $B$ moving from the inductive hypothesis to the inductive step because $\big\{I_{k}\big\}_{k=1}^{n+1}$ is an arbitrary finite collection of open intervals that covers $B$.
Please note that I recalled the proof using the fact that $[0,1]=\overline{B}\subset{\displaystyle{\overline{\bigcup_{k=1}^{n}I_{k}}=\bigcup_{k=1}^{n}\overline{I_{k}}}}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$ and since we are given a finite collection of open intervals, and then using finite subadditivity of $m^{*}$ afterwards -- posting that solution won't be necessary even though my proof, if it is correct, is more convoluted. Maybe their's a shorter way using induction even? I suppose I'm curious if above is correct since it then would be an alternate proof. Overall, my thanks goes out to any time spent on verifying my work.
You can proceed as you recalled. We have that
$$B \subset \overline B =[0,1] \subset \overline{\cup_{k=1}^n I_k} = \cup_{k=1}^n\overline{I_k}$$
which yields the following inequalities
$$m^*(B)\leqslant m^*([0,1]) = 1\leqslant m^*\left(\cup_{k=1}^n\overline{I_k}\right)$$
by monotonicity of the outer measure. Now, by (countable) subadditivity we have
$$m^*\left(\cup_{k=1}^n\overline{I_k}\right) \leqslant \sum_{k=1}^n m^*\left(\overline{I_k}\right),$$
and at this point it's enough to note that for any interval $I$ we have $m^*\left(I\right) = m^*\left(\overline{I}\right)$.
As a side note, this should also make it clear where the proof fails when the cover is allowed to contain countably many intervals, even though we have countable subadditivity. Can you spot it?