Trouble Following Proof of Heine-Borel Thorem using Nested Closed Intervals I am using the 2nd edition of S. Abbott's Understanding Analysis. The Heine-Borel theorem is laid out like so.

Theorem 3.3.8 (Heine–Borel Theorem). Let K be a subset of R. All of
the following statements are equivalent in the sense that any one of them implies
the two others: (i) K is compact. (ii) K is closed and bounded. (iii) Every open cover for K has a finite subcover.

The textbook defines a set to be compact if for every sequence in the set there exists a subsequence converging to a limit in the set.
In the book, it shows that the statements (i) and (ii) are equivalent. The book also walks through a proof that (iii) implies (ii) (and therefore (i)). The book leaves as an excercise to prove that (ii) implies (iii), the final piece in the puzzle of the proof.
I am having trouble with that latter part. It's a bit of a "guided proof". The question which asks for the proof of the Heine-Borel theorem is structured like so:

Exercise 3.3.9. Follow these steps to prove the final implication in Theorem
3.3.8.
Assume K satisfies (i) and (ii), and let $\{O_λ : λ ∈ Λ\}$ be an open cover for
K. For contradiction, let’s assume that no finite subcover exists. Let $I_0$ be a
closed interval containing K.

(a) Show that there exists a nested sequence of closed intervals $ I_0 ⊇ I_1 ⊇ I_2 ⊇
\dots $  with the property that, for each $n$, $I_n ∩ K$ cannot be finitely covered, and $\lim|I_n| = 0$ where $|I_n|$ is the length of the interval.
(b) Argue that there exists an $x ∈ K$ such that $x ∈ I_n$ for all $n$.
(c) Because $x ∈ K$, there must exist an open set $O_{\lambda_{0}}$  from the original collection that contains $x$ as an element. Explain how this leads to the desired
contradiction.
I think I'm just sort of lost on how to approach this, and any attempt I've made has pretty much fallen apart. Could anyone be kind enough to give a bit of a hint in the right direction, further than what the questions provide?
Thanks!
 A: Here's how you can proceed: $K$ is contained in $I_0$. Split $I_0$ into two halves. If the part of $K$ in both halves can be covered by finitely many open sets, then all of $K$ can be covered by combining these finitely many open sets. So, one of the halves must not be able to be covered by finitely many. Call that one $I_1$. (If neither can be covered, either will do, or you can decide to always take the left one.) Note that the length of $I_1$ is 1/2$|I_0|$.
Continuing to repeat this process, splitting the interval and picking the one where its intersection with $K$ can't be finitely covered, gives you the sequence of $I_n$, and their lengths must go to 0 since you keep dividing the length by 2 each time.
Then you'll need to use the compactness of $K$ in 2, and for 3 use that the $x$ is covered by some $O_\lambda$, $x$ is in every $I_n$, and the $I_n$ get arbitrarily small.
A: FYI. Here is the more common (and maybe the original) way to prove that (ii)$\implies$ (iii). The cases $K=\emptyset$ or $\min K=\max K$ are trivial so let's ignore them.
Let $\min K= k_0<k_1=\max K.$
Let $C$ be an open cover of $K$. Let $S\subset (k_0,4+k_1)$ where $r\in S$ iff  [$k_0<r< 4+k_1$ and $[k_0,r]\cap K$ can be covered by finitely many members of $C$]....  The "$4+k_1$" is just some arbitrary value greater than $k_1.$
$(\bullet)$. The goal is to show $k_1\in S.$
(i). There exists $c\in C$ and $r\in (k_0,4+k_1)$ such that $k_0\in [k_0,r]\subset c,$ so $S\ne \emptyset.$
(ii). If $k_0<r''<r\in S$ then $r''\in S.$ Because if $D$ is a finite subset of $C$ and $D$ covers $[k_0,r]\cap K$ then $D$ also covers $[k_0,r'']\cap K.$
(iii). We show that $\sup S>k_1$ by showing that if $k_1\ge r\in S$ then there exists $r'\in S$ with $r'>r$. Suppose $k_1\ge r\in S.$ Then:
Case (a): If $r\in K$: There exists $c\in C$ and $r'\in (r,4+k_1)$ such that $[r,r']\subset c,$ and there also exists a finite $D\subset C$ that  covers $[k_0,r]\cap K$ (because $r\in S$), so $D\cup \{c\}$ is a finite subset of $C$ that covers $[k_0,r']\cap K.$ So $r<r'\in S.$
Case (b): If $r\not\in K$: There exists $r'', r'$ with $k_0<r''<r<r' <4+k_1$ and $(r'',r']\cap K=\emptyset$ because $K$ is closed and $r\not \in K.$ Now $k_0<r''<r\in S$ so $r''\in S$ by (ii). So there exists a finite $D\subset C$ that covers $[k_0,r'']\cap K.$ And $D$ also covers $[k_0,r']\cap K$ because $(r'',r']\cap K=\emptyset.$ So $r<r'\in S.$
(iv). Since $\sup S>k_1$ there exists $t$ with $k_0<k_1<t\in S$ so $k_1\in S$ by (ii).
