Why do we need step 4. in this proof of the compactness of $[0, 1]$?

Question

In this proof of the compactness of $[0, 1]$ provided in this question

I am understanding proof of theorem stated in title from Spivak's calculus. It is as below.

(0) Let $\mathcal{O}$ be an open cover of $[0,1]$.

(1) Let $A=\{x\in [0,1]:[0,x] \mbox{ has finite subcover from } \mathcal{O}\}$.

(2) Then $A$ is non-empty, bounded above by $1$; let $\alpha$ be its supremum.

(3) Since $\mathcal{O}$ is open cover of $[0,1]$, $\alpha$ is in some $U$ from $\mathcal{O}$.

(4) There is an open interval $J$, $\alpha\in J\subseteq U$ s.t.all points of $J$ to the left of $\alpha$ are also in $U$.

(5) Since $\alpha$ is supremum of $A$, there is an $x\in J$ such that $x\in A$. How?

(6) Then $[0,x]$ is covered by finite subcover; this together with $U$ covers $[0,\alpha]$; so $\alpha\in A$.

(7) One tries to prove that $\alpha=1$, and proof will complete.

Q.1 It is in step 5, which I don't understand.

Q.2 Are there different proofs of this theorem? (I don't find other in 5-6 > standard books than this).

One of the answers to this question, by @Gribouillis has provided this explanation.

Q1

Since $\alpha = \sup(A)$, for any $\epsilon > 0$, there is an $x\in A$ such that $\alpha - \epsilon < x\le \alpha$. By choosing $\epsilon$ small enough, one has $(\alpha -\epsilon, \alpha]\subset J$, which proves (5).

Which was very helpful but has prompted me to ask what the purpose of line 4 in the proof is. First of all, it is quite redundant and doesn't add any restrictions to $J$ really. I mean, all points of $J$ to the left of $\alpha$ are also in $U$, which is obvious since all points of $J$ are in $U$ to begin with. One could just rewrite step 5. to use $U$ instead of $J$.

On the other hand, because $J$ might be made out of multiple disconnected open intervals, the fact that $\alpha$ is an element of $J$ along with another element $x$ from $A$ doesn't automatically mean the finite open covering of $[0,x]$ along with $J$ covers $[0, \alpha]$. It would be more meaningful to assert that $J$ is a single interval in order to cover $[0, \alpha]$

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Edit: I have misread $J$ as an open subset of $U$ rather than an open interval which it is. As a result, I mistakenly question the need for step 4. based on reading it as

(4) There is an open set $J$, $a \in J \subseteq U$ ...

rather than

(4) There is an open interval $J$, $a \in J \subseteq U$ ...

Answer 1

The purpose of step 4 is to provide an interval which contains $\alpha$ and some point $x \in A$. As you say, $U$ may be made of disconnected intervals, so we need to first pick an interval within $U$ containing $\alpha$. The line about it containing all points of $J$ to the left of $\alpha$ is superfluous.

Alternatively, you could first refine $\mathcal{O}$ by breaking each open set into intervals, skipping step 4 entirely.

Answer 2

In your question you restate Spivak's proof. It seems that you already found the answer to Q1. However, your formulation of $(4)$ is misleading:

There is an open interval $J$, $\alpha\in J\subseteq U$ s.t.all points of $J$ to the left of $\alpha$ are also in $U$.

This suggests that the "such that" part is an additional requirement on $J$, but it is just a consequence of the fact that $J \subseteq U$. Anyway, $\alpha \in J \subseteq U$ is all we need to know, and this is an application of the definition of open subsets in $\mathbb R$. Using the definition is of course indispensable.

Writing $J = (a,b)$, we see that $a < \alpha$. Since $\alpha = \sup A$, there exist $x \in A$ such that $a < x \le \alpha$. Then $[x,\alpha] \subset J \subset U$. Since $[0,x]$ is covered by finitely many elements of $\mathcal O$, we are done.

Concerning Q2: Spivak's proof is the standard proof. It is completely elementary and very short, thus I do not see the need to find alternatives.