TITLE: Showing that $[0,1]$ is compact
QUESTION [8 upvotes]: Let's choose an open covering for $\left [ 0,1 \right ]$. For example $$\left \{ \left ( \frac 1 n,1-\frac 1 n \right )  \mid  n\in \{ 3,4,\dots\} \right \}.$$
How can one choose a finite open subcover to prove compactness?

REPLY [49 votes]: Here is a quick and elegant proof of the actual result that $[0,1]$ is compact based on real induction:
Let $\mathcal{O}$ be an (arbitrary!) open cover. Let $P$ be the set of points $x$ in $[0,1]$ such that $[0,x]$ can be covered by finitely many elements of $\mathcal{O}$. We have $0\in P$ and $P$ is bounded above by $1$. Therefore, $P$ has a supremum $s$. 
We first show that $[0,s]$ can be covered by finitely many sets in $\mathcal{O}$. This is trivial when $s=0$, so assume $s>0$. Let $O_s\in\mathcal{O}$ be a set containing $s$. Then there is an $\epsilon \in (0, s)$ such that $(s-\epsilon,s]\subseteq O_s$. By assumption, there is a finite subcover of $[0,s-\epsilon/2]$. By adding $O_s$ to that finite subcovering, we get a finite subcovering of $[0,s]$.
We now show that $s=1$. Suppose $s<1$ and let $O_s\in\mathcal{O}$ be a set containing $s$. Then there is an $\epsilon>0$ such that $[s,s+\epsilon)\subseteq O_s$. So taking a finite subcover of $[0,s]$ and adding the set $O_s$ gives us a finite subcover of $[0,s+\epsilon/2]$, contradicting the construction of $s$.