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

Write $A=[a,b]$, and let $\mathcal{T}_{A}$ be the subspace topology on $A$ induced by $\mathcal{T}_{\mathbb{R}}$. By Compact Topological Space and Compact Subset §compact-subset and Compact Topological Space and Compact Subset §compact we must show that for every subset $\mathcal{U}$ of $\mathcal{T}_{A}$ with $A\subseteq\bigcup\mathcal{U}$ there is a finite subset of $\mathcal{U}$ whose union contains $A$. Fix such a set $\mathcal{U}$. The order of $\mathbb{R}$ is total, and the rules of Rules of Arithmetic and Order in an Ordered Field apply to it.

Step 1 (the set of good right endpoints). For $t\in A$ put $A_{t}=\{u\in A:u\le t\}$, and let $S$ be the set of all $t\in A$ for which there is a finite subset $\mathcal{V}$ of $\mathcal{U}$ with $A_{t}\subseteq\bigcup\mathcal{V}$. Since $a\le a\le b$, we have $a\in A$; and $A_{a}=\{a\}$, as $u\in A_{a}$ means $a\le u\le a$. As $A\subseteq\bigcup\mathcal{U}$, there is $U_{0}\in\mathcal{U}$ with $a\in U_{0}$, and $\{U_{0}\}$ is a finite subset of $\mathcal{U}$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small with $A_{a}\subseteq\bigcup\{U_{0}\}$. Hence $a\in S$, so $S$ is a nonempty subset of $\mathbb{R}$, and $b$ is an upper bound of $S$ because $S\subseteq A$. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum, $S$ has a supremum, unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum; call it $s$. Then $a\le s$ as $a\in S$, and $s\le b$ as $s$ is the least upper bound and $b$ an upper bound; so $s\in A$.

Step 2 (an open set around $s$). Since $s\in A\subseteq\bigcup\mathcal{U}$, there is $U\in\mathcal{U}$ with $s\in U$. As $U\in\mathcal{T}_{A}$, Subspace Topology §subspace-topology gives $W\in\mathcal{T}_{\mathbb{R}}$ with $U=A\cap W$. Then $W$ is open in $(\mathbb{R},d_{\mathbb{R}})$ and $s\in W$, so there is a real number $r>0$ with $B_{d_{\mathbb{R}}}(s,r)\subseteq W$, the open ball of centre $s$ and radius $r$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, adding $s-r$ to $0<r$ gives $s-r<s$, so $s-r$ is not an upper bound of $S$, as $s$ is the least one; hence there is $t\in S$ with $s-r<t$, the order being total. Fix such a $t$ and a finite subset $\mathcal{V}$ of $\mathcal{U}$ with $A_{t}\subseteq\bigcup\mathcal{V}$, and put $\mathcal{V}'=\mathcal{V}\cup\{U\}$, a finite subset of $\mathcal{U}$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union.

Step 3 (every $u\in A$ with $u<s+r$ lies in $\bigcup\mathcal{V}'$). Let $u\in A$ with $u<s+r$. If $u\le t$, then $u\in A_{t}\subseteq\bigcup\mathcal{V}$. Otherwise $t<u$, so $s-r<u$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $s-r<u$ gives $s-u<r$ and $u<s+r$ gives $u-s<r$. By Absolute Value in an Ordered Field §absolute-value and Rules of Arithmetic and Order in an Ordered Field §signs, $|s-u|$ is $s-u$ or $-(s-u)=u-s$, so $|s-u|<r$, that is $d_{\mathbb{R}}(s,u)<r$ by The Distance on the Real Numbers §distance. By Open Ball in a Metric Space §membership, $u\in B_{d_{\mathbb{R}}}(s,r)\subseteq W$, and as $u\in A$ we get $u\in A\cap W=U$, so $u\in\bigcup\mathcal{V}'$.

Step 4 ($b\in S$). Since $s<s+r$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, Rules of Arithmetic and Order in an Ordered Field §midpoint gives $s<m<s+r$ for $m=(s+(s+r))/(1+1)$. Let $t'=\min\{m,b\}$, which exists by The Maximum and Minimum of Two Elements of a Total Order §exists. By The Maximum and Minimum of Two Elements of a Total Order §bounds, $t'\le m$ and $t'\le b$; and as $a\le s<m$ and $a\le b$, The Maximum and Minimum of Two Elements of a Total Order §greatest-lower gives $a\le t'$. So $t'\in A$, and every $u\in A_{t'}$ satisfies $u\le t'\le m<s+r$, hence $u\in\bigcup\mathcal{V}'$ by Step 3. Thus $A_{t'}\subseteq\bigcup\mathcal{V}'$, so $t'\in S$ and $t'\le s$. Since $s<m$, $t'\neq m$. If $m\le b$, then $t'=m$ by The Maximum and Minimum of Two Elements of a Total Order §values, which is impossible; so $b\le m$, the order being total, and $t'=b$ by The Maximum and Minimum of Two Elements of a Total Order §values.

Hence $A_{b}\subseteq\bigcup\mathcal{V}'$. Every $u\in A$ satisfies $u\le b$, so $A_{b}=A$, and $\mathcal{V}'$ is a finite subset of $\mathcal{U}$ with $A\subseteq\bigcup\mathcal{V}'$. As $\mathcal{U}$ was arbitrary, $(A,\mathcal{T}_{A})$ is compact, that is, $[a,b]$ is compact in $(\mathbb{R},\mathcal{T}_{\mathbb{R}})$.