Proof of Closed Interval $[a,b]$ is Compact in $\mathbb{R}$

By Compact Subset Criterion via Open Covers in the Ambient Space, it is enough to prove that every open cover of $[a,b]$ in $\mathbb{R}$ has a finite subcover.

If $a=b$, then $[a,b]=\{a\}$. Let $(U_i)_{i\in I}$ be an open cover of $[a,b]$ in $\mathbb{R}$. Since $a\in[a,b]$, there exists $i_0\in I$ with $a\in U_{i_0}$. Then the single set $U_{i_0}$ already covers $[a,b]$. So $[a,b]$ is compact in this case.

Assume from now on that $a<b$, and let $(U_i)_{i\in I}$ be an open cover of $[a,b]$ in $\mathbb{R}$. Define

$$S=\{x\in[a,b]: [a,x] \text{ has a finite subcover drawn from } (U_i)_{i\in I}\}.$$

Since $a\in[a,b]$, there exists $i_0\in I$ with $a\in U_{i_0}$. Because $U_{i_0}$ is open in the Euclidean sense, Open Subset of Euclidean Space gives a real number $r>0$ such that every $y\in\mathbb{R}$ satisfying

$$(y-a)^2<r^2$$

belongs to $U_{i_0}$. If $x\in[a,b]$ and $a\le x<a+r$, then $(x-a)^2<r^2$, so $x\in U_{i_0}$. Since $a<b$, there exists $c\in[a,b]$ with $c>a$ and $c<a+r$; for example one may take $c=\min\{b,a+r/2\}$. Then $[a,c]\subseteq U_{i_0}$. Therefore $c\in S$, so $S$ is nonempty.

Also $S\subseteq[a,b]$, so $S$ is bounded above by $b$. By the least upper bound property, there exists a real number $s\in\mathbb{R}$ such that $s=\sup S$ in the sense of the supremum definition.

We claim that $s=b$. Suppose instead that $s<b$. Since $s\in[a,b]$, the cover property gives some $j\in I$ with $s\in U_j$. Because $U_j$ is open in the Euclidean sense, there exists $r>0$ such that every $y\in\mathbb{R}$ with

$$(y-s)^2<r^2$$

belongs to $U_j$.

By the approximation property in Supremum (least upper bound), applied with $\varepsilon=r/2$, there exists $t\in S$ such that

$$s-r/2<t\le s.$$

Because $t\in S$, there are finitely many members of the cover whose union contains $[a,t]$. We show that the same finite family together with $U_j$ covers a larger interval. Let $x\in[a,s+r/2]$. If $x\le t$, then $x$ lies in one of the finitely many sets already covering $[a,t]$. If $x\ge t$, then

$$0\le x-s\le r/2, \qquad 0\le s-x<r,$$

so in either case $(x-s)^2<r^2$. Hence $x\in U_j$. Therefore $[a,s+r/2]$ has a finite subcover drawn from $(U_i)_{i\in I}$, so $s+r/2\in S$. This contradicts that $s$ is an upper bound for $S$.

Thus $s=b$. Since $b=\sup S$, one has $b\in S$. Hence $[a,b]$ has a finite subcover. By Compact Subset Criterion via Open Covers in the Ambient Space, the interval $[a,b]$ is compact in $\mathbb{R}$.