TITLE: How to prove every closed interval in R is compact?
QUESTION [64 upvotes]: Let $[a,b]\subseteq \mathbb R$. As we know, it is compact. This is a very important result. However, the proof for the result may be not familar to us. Here I want to collect the ways to prove $[a,b]$ is compact. 
Thanks for your help and any link.

REPLY [4 votes]: We have that $a<b$.
Take an open cover
\begin{align}
\left[ a,b \right] \subset \bigcup_{\alpha \in A}U_{\alpha}.
\end{align}
 Let $S$ be the set of all $x\in\left[ a,b\right]$ such that $\left[ a,x\right]$ is covered by finitely many $U_\alpha$. We want to show that $b$ is in this set $S$.
Clearly there is some $\alpha_0$ such that $a\in U_{\alpha_0}$ so we know that $S$ is a non-empty subset of $\mathbb{R}$ bounded above by $b$. We want to show that 
 \begin{align*}
 x_0=\sup S \in [a,b]=b.
 \end{align*}
We assume the contrary, that $x_0<b$. We also have that $x_0>a$ since there is $\epsilon>0$ and some $\alpha_0$ such that $\left[ a,a+\epsilon \right] \in U_{\alpha_0}$, so $x\geq a+\epsilon$.
We then take $\alpha_0$ such that $x_0\in U_{\alpha_0}$ and take $\epsilon>0$ such that $a\leq x_0-\epsilon<x_0<x_0+\epsilon\leq b$ and such that $\left[ x_0-\epsilon,x_0+\epsilon\right] \subset U_{\alpha_o}$.
Since $x_0-\epsilon$ is not an upper bound there exists $x_1 \in S $ where $x_0-\epsilon \leq x_1 \leq x_0$. But then $[a,x_1]$ is finitely covered by $U_\alpha$'s,
\begin{align*}
\left[ a,x_1\right] \subset \bigcup_{\beta=1}^n U_{\alpha_\beta}
\end{align*}
which implies
\begin{align*}
\left[ a,x_0+\epsilon \right] \subset \bigcup_{\beta=1}^n U_{\alpha_\beta}\cup U_{\alpha_0}.
\end{align*}
This implies $x_0+\epsilon \in S$ contradicting that $x$ is the upper bound. We get then that $\sup S=b$ and similar argument shows that $b\in S$. In fact we have we have $[a-\epsilon,b]\subset U_{\alpha_0}$ for some $\epsilon>0$ and some $\alpha_0$. There is then some $x_1$ in  $[a-\epsilon,b]$ belonging to $S$. Then there is a finite subcover for $[a,b]=[a,x_1]\cup [x_1,b]$. The segment $[a,b]$ is then compact as expected.