Prove that a full cover of an interval [a, b] has a finite partition The following is an exercise from Bruckner's Real Analysis:

I only have a problem with item (d). Of course one can use Heine–Borel theorem first by contracting all intervals by scale 9/10 then remove end points to form an open covering then finding a finite cover removing interior of intersection of intervals (using definition of full cover), then adding the end points to solve the item (d). But the exercise wants to use the hint especially when later in the book asks to prove the Heine–Borel theorem based on item (d). I have no idea how to use the hint to solve the exercise. Please help, thanks!
Definition : Let $\mathcal{I}$ be the family of nondegenerate closed intervals in $\mathbb{R}$. Let $E \subset \mathbb{R}$ and let $\mathcal{V} \subset \mathcal{I}$. If  for each $x \in E$ and $ε > 0$ there exists $V \in \mathcal{V}$ such that $x \in V$ and $λ(V) < ε$, then $\mathcal{V}$ is called a Vitali cover for $E$. The exercise defines full cover analogously.
 A: Let us insist that all partitions we mention are finite and that the intervals can overlap at endpoints as per the book. Note it is impossible to disjointly cover an interval with finitely many closed intervals.
Let $A$ be the set considered in the Hint. We need to prove to claims. 1) For every element $x\in A$ with $x<b$ there is $y\in A,x<y$ 2) $\sup{A} \in A$

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*For $x\in A$ we can find $\epsilon>0$ such that $[x,x+\epsilon]$ is in the full cover. Given a partition of $[a,x]$ we can add $[x,x+\epsilon]$ to it to get a partition of $[a,x+\epsilon].$


*There exists $\epsilon >0$ such that $\forall \delta < \epsilon, [\sup{A}-\delta,\sup{A}]\in W.$ Pick $x\in A$ such that $\sup{A}-x<\epsilon.$ Then, $[a,x]$ has a partition which we can extend with $[x,\sup{A}].$ Hence $\sup{A}\in A.$
Together 1) and 2) tell us that $\sup{A}=b$ and hence $[a,b]$ has a partition as desired.
A: Answering item (d). We can use a simpler hint.
Let $\mathcal{W}$ be a full cover of $[a,b]$.
Let $A = \{ x\in (a,b]: \text{ there is a finite partition of } [a, x] \text{ using members of } \mathcal{W} \}$.

*

*$A \ne \emptyset$. In fact, let $[a,c]$ (with $c>a$) be a sufficiently small interval containing $a$ so that
$[a,c] \in \mathcal{W}$. Then $c \in A$, because $[a,c]$ is a trivial finite partition of $[a,c]$.


*$\sup A \in A$. Let $s= \sup A$ and let $\{x_n\}_n$ be an increasing sequence of elements in $A$ such that $x_n \nearrow s$. Then, there $k$ such that $(x_k, s]$ is a sufficiently small interval containing $s$ so that
$(x_k,s] \in \mathcal{W}$.
Since $x_k \in A$,  there is a finite partition of $ [a, x_k] $ using members of $\mathcal{W}$. Adding $(x_k,s]$ to such partition, we have a finite partition of $ [a, s] $ using members of $\mathcal{W}$. So $s \in A$.


*$\sup A = b$. Let $s= \sup A$. Clealy $s \leq b$. Suppose $s<b$. Then there is $d$, such that $s<d<b$ and $[s, d]$ is a sufficiently small interval containing $s$ so that
$[s,d] \in \mathcal{W}$. Since $s \in A$,  there is a finite partition of $ [a, s] $ using members of $\mathcal{W}$. Adding $[s,d]$ to such partition, we have a finite partition of $ [a, d] $ using members of $\mathcal{W}$. So $d \in A$. Contradiction because $s = \sup A$ and $s< d$. So we must have $s=b$, that is $\sup A = b$.
So $b \in A$, which means that there is a  finite partition of $ [a, b] $ using members of $\mathcal{W}$.
Remark: When forming a partition, it is allowed that the intervals overlap in the extreme points. And this is used in 3.
