How to prove every closed interval in R is compact? 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 familiar to us. Here I want to collect the ways to prove $[a,b]$ is compact.
Thanks for your help and any link.

Here is a different proof.  Alexander's lemma.  (It states: given a subbase for the topology, $X$ is compact iff every cover by elements of the subbase has a finite subcover.)  Now use the sets
$$
[0,a), 0<a<1,\qquad (b,1], 0<b<1
$$
as your subbase $\mathcal B$.
Suppose $\mathcal A \subseteq \mathcal B$ is a cover of $[0,1]$.  Since $1 \in [0,1]$, there is at least one set of the form $(b,1]$ in $\mathcal A$. Let $b_0 = \inf\{b : (b,1] \in \mathcal A\}$.  Then, since $b_0 \in [0,1]$ there is $[0,a_1) \in \mathcal A$ such that $b_0 \in [0,a_1)$.  And since $a_1>b_0$, by definition of inf there is $(b_1,1] \in \mathcal A$ with $b_1<a_1$.  So in fact $[0,1] = [0,a_1) \cup (b_1,1]$.  Not only is there a finite subcover, there is a subcover of just two sets.
