Every open cover of $[0,1]$ has finite subcover 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).
 A: Q2
Suppose the contrary, that is, assume that there is no finite subcover that covers $[0,1]$. Then, at least one of the intervals $[0,1/2]$ and $[1/2, 1]$ can't be covered with a finite subcover. Pick this interval and call it $I_1$. This is the base of an inductive process.
Now assume that $I_n$ has length $1/2^n$ and can't be covered with a finite subcover. Divide $I_n$ into two intervals of the same length. One of these intervals can't be covered with a finite subcover. Pick this one and call it $I_{n+1}$ This new interval has length $1/2^{n+1}$. This completes the inductive process, which yields a sequence of intervals $I_n$. Each interval of the sequence is contained and has a length that is a half of the preceeding one. And none of them can be covered with a finite subcover.
Since $I_{n+1}\subset I_n$, $\bigcap_{n=1}^\infty I_n$ is not empty. And since the length of $I_n$ tends to $0$ it has only one point. That point is in some open set from the cover, and this open set alone covers some $I_m$. Contradiction.
Remark: this idea can be used to show that a finite cartesian product of closed, bounded invervals is compact.
A: 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).
A: A set $A$ is sequentially compact if every $x_n \in A$ has a convergent subsequence.

$Theorem:$ Every sequentially compact metric space $X$ is compact.

For a proof see this:
https://people.clas.ufl.edu/mjury/files/sequential_compactness_notes.pdf
Now let $x_n \in [0,1] \Rightarrow x_n$ is bounded thus from Bolzano-Weierstrass theorem is has a convergent subsequence therefore $[0,1]$ is sequentially compact.
From the above theorem $[0,1]$ is compact.
A: From a general topology point of view, you could also apply Alexander's subbase lemma (see my answer on Topology Atlas here where the final application is the fact that in ordered space $X$, $X$ is compact iff every subset of $X$ has a supremum. So that's a different proof, but it also relies (as it must) on the completeness of the reals.
You could also prove it's complete and totally bounded (a uniform space approach).
