Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction?
Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{F}$ containing $x$ with $\ell (I) < \epsilon$. I tried choosing $\epsilon > 0$ small enough so that it $I$ wouldn't intersect any of the $I_k$, thereby showing $I \in \mathcal{F}_n$; but it wasn't clear to me how to do this. Indeed, it doesn't seem possible...[Note: the errata sheet for Royden-Fitpatrick's Real Analysis says that $\infty$ should be replaced by $n$]
Where are we getting all of the disjoint collections? What guarantees they exist? E.g., "Suppose $n$ is a natural number and the finite disjoint subcollection $\{I_k\}_{k=1}^n$ has been chosen." How has this been chosen? It seems that we've chosen them out of thin air.
I could keep going on. At this point I'm pretty lost given the sheer number of choices he has made and will make (e.g., why can we choose $\ell(I_{n+1}) > s_n/2$?). It isn't terribly clear how $\{I_k\}_{k=1}^n$ is obtained and how the rest are obtained.
EDIT:
I with the help of Tony S.F., I have been able to resolve the 1st and 2nd parts of my question. For the 1st, given $x \in E \subseteq \mathcal{O}$, there is some $r > 0$ such that $B(x,r) \subseteq \mathcal{O}$. Given $\frac{r}{2} > 0$, there is a compact interval $I = [a,b] \in \mathcal{F}$ with $\ell (I) < \frac{r}{2}$ such that $x \in I$. Hence $b-a < \frac{r}{2}$ or $b < \frac{r}{2} + a$, and $a \le x \le b$. From these we get $a \le x < a + \frac{r}{2}$ or $|x-a| < \frac{r}{2}$. Hence, if $y \in I = [a,b]$, then
\begin{align} |x-y| &= |(x-a) + (a-y)| \\ &\le |x-a| + |y-a| \\ &< \frac{r}{2} + \frac{r}{2} = r, \\ \end{align} and therefore $y \in B(x,r)$, from which it follows $I \subseteq B(x,r) \subseteq \mathcal{O}$.
Now for the second part. Let $\{I_k\}_{k=1}^n \subseteq \mathcal{F}$ and let $x \in E- \bigcup_{k=1}^n$. Then $x \notin I_k$ for every $k$, and, as the $I_k$ are closed, there exists $r_k > 0$ for which $B(x,r_k) \cap I_k = \emptyset$. Letting $r = \frac{1}{2} \min\{r_1,...,r_n\}$, there exists $I_r$ with $\ell(I_r) < r \le \frac{r_k}{2}$ for each $k$ such that $x \in I_r$. But, as we saw above, this means $I_r \subseteq B(x,r_k)$ for each $k$ and therefore $I_r \cap I_k = \emptyset$ for each $k$. Hence $x \in I_r \subseteq \bigcup_{I \in \mathcal{F}_n} I$.
Point 3 is still giving me trouble. I still don't understand this: "Suppose $n$ is a natural number and the finite disjoint subcollection $\{I_k\}_{k=1}^n$ has been chosen." What justifies this supposition?
Granting that for a moment, I think I now see how $I_{n+1}$ is chosen. Since $s_n := \sup \{\ell (I) \mid I \in \mathcal{F}_n\}$ is finite, given $s_n/2 > 0$, there exists $I \in \mathcal{F}_n$ for which $s_n < \ell(I) + s_n/2$ or $\ell(I) > s_n/2$. Let $I_{n+1}$ equal this particular $I \in \mathcal{F}_n$. The only thing I don't see is why $\ell (I_{n+1}) > \ell(I)/2$ for every $I \in \mathcal{F}$. Certainly $\ell (I_{n+1}) > \ell(I)/2$ for every $I \in \mathcal{F}_n$ is true, because by definition $\ell(I_{n+1}) > s_n/2 \ge \ell(I)/2$ for every $I \in \mathcal{F}_n$.
The last thing giving me trouble is how $I \cap I_k = \emptyset$ for every $k$ implies $\ell(I_k) > \ell(I)/2$ for every $k$.
If I have followed the edits and comments correctly, then there are two remaining aspects of the proof that you are confused about.
First, there is point 3. To understand what's going on here, you need to appreciate that the sequence $\{I_k\}_{k=1}^\infty$ is being defined inductively. A definition by induction works very much like a proof by induction, and I assume you have some familiarity with the latter. One wants to define, for every $n \in \mathbb{N}$, an interval $I_n$ such that the members of $\{I_k\}_{k=1}^n$ are pairwise disjoint and $\cup_{k=1}^n I_k$ satisfies (5). One begins by defining a base case $I_1$, which, here, is an arbitrary interval. One then proceeds with the inductive part of the definition: assuming that the first $n$ intervals have been appropriately defined, define $I_{n+1}$. The two steps of the definition---the base case and the inductive step---ensure that the entire sequence $\{I_k\}_{k=1}^\infty$ is well-defined.
Another way to describe what's going on here is to think of the procedure for choosing the $\{I_k\}_{k=1}^\infty$ as follows. Choose $I_1$ arbitrarily. If $E - I_1 = \emptyset$, then the proof is complete. Otherwise, choose any $I_2 \in \mathcal{F}_1$ such that $\ell(I_2) > s_1/2$. If $E - (I_1 \cup I_2) = \emptyset$, then the proof is complete. Otherwise, choose any $I_3 \in \mathcal{F}_2$ such that $\ell(I_3) > s_2/2$. And so on up to $I_n$. At each stage $n$, either the proof is completed or $I_n$ is well-defined. Thus, assuming that the proof isn't completed after choosing finitely many intervals, the entire sequence $\{I_k\}_{k=1}^\infty$ is well-defined.
The second remaining problem is: Why does $I \cap I_k = \emptyset$ for all $k$ imply $\ell(I_k) > \ell(I)/2$ for all $k$?
If $I \cap I_k = \emptyset$ for all $k$, then $I \cap (\cup_{k=1}^n I_k) = \emptyset$ for all $n$. Then (6) implies that $\ell(I_{n+1})>\ell(I)/2$ for all $n$, which, by changing the indexing variable, is the same as saying $\ell(I_k) > \ell(I)/2$ for all $k$.