Show that there is a countable disjoint collection $\{ I_k \}_{k = 1}^{\infty}$ of intervals

To all who are reading this: please also read the answer by the other user. An issue with my answer that I didn’t realise was raised in the comment section and fixed in the other answer.

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We construct $\{I_k\}_{k=1}^{\infty}$ inductively.

Enumerate the natural numbers with $m$, beginning with $m=1$.

Let $\epsilon=\frac{1}{m}$.

Find inductively, using the Vitali Covering Lemma, a finite disjoint collection of closed intervals, $\{I_{mk}\}_{k=1}^{n_m}$, for which $$m^* (E_m) < \frac{1}{m},$$

where we define $E_1= E \sim \bigcup_{k =1}^{n_1} I_k$ and, for subsequent $m$'s, $E_m=E_{m-1} \sim \bigcup_{k =1}^{n_m} I_k.$

It follows that $m^* (E_m) < \frac{1}{m}$ is arbitrarily small. We let our $\{I_k\}_{k=1}^{\infty}$ to be the collection of all the intervals that we have found. There are $\sum_{m=1}^{\infty}n_m$ of them -- that is a countable collection of intervals.

We can make this collection disjoint, by, at the beginning of each round, taking out the ones that would intersect the intervals already chosen in the previous rounds.

This answer is for those who have access to the text mentioned in the question, Real Analysis, fourth edition [2018 reissue] by Royden and Fitzpatrick.

The disjoint countable subcollection $\{I_k\}_{k=1}^\infty$ that the authors construct in their proof of the Vitali covering lemma will do for this problem. Indeed, we deduce from (5) and the last paragraph of the proof that there is an index $N$ for which if $n\ge N$, then $$m^*\Biggl(E\sim\bigcup_{k=1}^n I_k\Biggr)\le\sum_{k=n+1}^\infty 5\ell(I_k) <\epsilon.$$ Thus, by the definition of convergence (page 21),\begin{align} m^*\Biggl(E\sim\bigcup_{k=1}^\infty I_k\Biggr) & =\lim_{n\to\infty} m^*\Biggl(E\sim\bigcup_{k=1}^n I_k\Biggr)\\ &\le\lim_{n\to\infty}\sum_{k=n+1}^\infty 5\ell(I_k)\\ & = 0.\end{align}

I am including this answer which is based on Odysseus F's idea but without the issues I raised in my comments there. I find my other answer, however, more satisfying than this one.

References are to Real Analysis, fourth edition [2018 reissue] by Royden and Fitzpatrick.

The proof is similar to that of Theorem 11(ii) of Section 2.4. By the Vitali covering lemma and its notation, for each natural number $p$, there is a finite disjoint subcollection $\{I_{pi}\}_{i=1}^{n_p}$ of $\mathcal F$ for which $$m^*\Biggl(E\sim\bigcup_{i=1}^{n_p} I_{pi}\Biggr) <\frac1p.$$ Put all those $I_{pi}$ in a single countable collection $\bigl\{I'_j\bigr\}{}_{j=1}^\infty$. That collection is countable because it is the union of a countable (index $p$) collection of countable sets (each with $n_p$ intervals) (Corollary 6 of Section 1.3). Moreover, because for each $p$, $E\sim\bigcup_{j=1}^\infty I'_j\subseteq E\sim\bigcup_{i=1}^{n_p} I_{pi}$, by monotonicity of outer measure, $$m^*\Biggl(E\sim\bigcup_{j=1}^\infty I'_j\Biggr)\le m^*\Biggl(E\sim\bigcup_{i=1}^{n_p} I_{pi}\Biggr) <\frac1p.$$ Now if two intervals in $\bigl\{I'_j\bigr\}{}_{j=1}^\infty$ are not disjoint, discard the one associated with the smaller $p$ index to form a countable disjoint collection $\{I_k\}_{k=1}^\infty$ of intervals in $\mathcal F$. The above inequality still holds with the $I'_j$s replaced by the $I_k$s because as we step through the interval-discarding procedure, the collection of $I_k$s contains collection $\{I_{pi}\}_{i=1}^{n_p}$ at step $p$. Therefore, $$m^*\Biggl(E\sim\bigcup_{k=1}^\infty I_k\Biggr)= 0.$$