Help with understanding the proof of $[a,b]$ is compact. I am having trouble understanding certain parts of the proof of theorem 2 that my professor gave. Could someone kindly clarify these parts?
First, a theorem he used in the proof:
$\textbf{Theorem 1.}$ ($\textit{Nested Interval Theorem}$) If $\{I_n\}$ is a sequence of closed and bounded intervals in
$\mathbb{R}^1$
such that
$I_n \supseteq I_{n+1}, \forall n \geq 1$
then
$\bigcap\limits_{i=1}^{\infty} I_n \neq \emptyset$.
Now, the proof:
$\textbf{Theorem 2.}$ Every closed interval $[a,b]$ is compact.
$\textbf{Proof.}$ Let $I = \{x \ |\ a \leq x \leq b \}$ and let $\delta = b-a$. Then $|x-y| \leq  \delta$ if $x,y \in I$.
Suppose there exists an open covering $\mathscr{G} = \{G_\alpha\}$ of $I$ which has no finite covering. Let $c= \frac{a+b}{2}$, so 
$I = [a,c] \cup [c,b]$.
At least one of these intervals cannot be covered by any finite subcover of $\mathscr{G}$; call it $I_1$. We next bisect $I_1$ and continue this process. We get a sequence $\{I_n \}$ of closed intervals such that
$(i)\ I_n \supseteq I_{n+1} \forall n \geq 1$;
$(ii)\ I_n$ is $\textbf{not}$ covered by any finite subcollection of $\mathscr{G}$;
$(iii)$ If $x,y \in I_n$, then $|x-y| \leq \frac{\delta}{2^n}$.
By Theorem 1, $\exists\ x^* \in$ each $I_n$. So $x^* \in$ some $G_\alpha$.
$\textbf{I don't understand this next line. Why must there exist such an r?}$
Since $G_\alpha$ is open, $\exists\ r > 0$ such that
$$|y-x^*|<r\ \Rightarrow\ y \in G_\alpha.$$
$\textbf{I also don't understand this next sentence. I am not seeing how n being so large implies}$ $(iii) \Rightarrow I_n \subseteq G_\alpha.$
If $n$ is so large that $\frac{\delta}{2^n} < r$ (there is such an $n$, for otherwise $2^n \leq \frac{\delta}{r}\ \forall\ n \in \mathbb{Z}^+$, which is not possible becasue $\mathbb{R}$ is Archimedean), then $(iii)$ implies $I_n \subseteq G_\alpha$ which contradicts $(ii)$. $□$
Thank you for your help!
 A: I think there is a typo on the line after (iii). It should read $``$So $x^*\in$ some $G_\alpha$$"$ instead of $``$$x^*\in$ each $G_\alpha$$"$. 
Why must there exist such an $r$? This is because $x^*$ is an element of $G_\alpha$ which is an open set. So by definition there exists an open interval of some radius $r$ centered around $x^*$ and inside $G_\alpha$ i.e. $(x^*-r, x^*+r)\subset G_\alpha$ i.e., if $y\in (x^*-r, x^*+r)$ then $y\in G_\alpha$ and so $|y-x^*|<r\Rightarrow y\in G_\alpha$.
Now if $n$ is so large that $\frac{\delta}{2^n}<r$ then for $y\in I_n$, (by (iii)) $$|y-x^*|<\frac{\delta}{2^n}<r$$ which implies by above that $y\in G_\alpha$. So $I_n\subset G_\alpha$. This is a contradiction because by construction, $I_n$ is not supposed to be coverable by any finite subcollection of $\mathscr{G}$.
A: *

*That such an $r$ exists is directly from the definition of the metric space topology in $\mathbb{R}$: Any open set $U$ in the real line is a union of sets $B(x, r) = \{y \in \mathbb{R} \, \, ; |x - y| < r\}$ running over $x \in U$ and some collection of numbers $r > 0$.

*At some $n$ the set $I_n$ will be contained in the ball $B(x^*, r)$ because every $y \in I_n$ will be within $r$ of the point $x^*$ --- as described in the steps following your second question. In other words, every point $y \in I_n$ is a point of $B(x^*, r)$ which, using your previous assertion, is a point of $G_\alpha$.
That means some $I_n$ is contained in a single $G_\alpha$, which gives you the contradiction.
