Proof of Extreme Value Theorem on a Closed Real Interval

Proof of (ii). We apply Closed Interval $[a,b]$ is Compact in $\mathbb{R}$ (i) and Extreme Value Theorem on a Compact Subset of a Metric Space (i) to the metric space $(X,d)=(\mathbb{R},d_{\mathbb{R}})$. Both statements use this distance $d_{\mathbb{R}}$, and their topologies $\mathcal{T}_{\mathbb{R}}$ and $\mathcal{T}_{d}$ are then the same set $\mathcal{T}_{d_{\mathbb{R}}}$ of open subsets of $\mathbb{R}$, which exists by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii); and $[a,b]$, the set of all $t\in\mathbb{R}$ with $a\le t\le b$ by Intervals of the Real Line (ii), is the set $[a,b]$ of the first statement. So the first statement, applied to $a\le b$, shows that $K=[a,b]$ is compact in $(\mathbb{R},\mathcal{T}_{d_{\mathbb{R}}})$. Moreover $K\neq\emptyset$, since $a\le a\le b$ gives $a\in K$, and $f:K\to\mathbb{R}$ is continuous on $K$ by hypothesis. Thus the second statement applies to $K$ and $f$, and gives $x_{\min},x_{\max}\in[a,b]$ with $f(x_{\min})\le f(x)\le f(x_{\max})$ for every $x\in[a,b]$. Put $p=x_{\min}$ and $q=x_{\max}$.

Proof of (i). Let $p$ and $q$ be as in (ii), and put $M=|f(p)|+|f(q)|$, a real number. Let $x\in[a,b]$. By Rules of Arithmetic and Order in an Ordered Field (x), $-|f(p)|\le f(p)$ and $f(q)\le|f(q)|$, and $|u|\le M$ for every real $u$ with $-M\le u\le M$. As $0\le|f(p)|$ and $0\le|f(q)|$,

$$-M\le-|f(p)|\le f(p)\le f(x)\le f(q)\le|f(q)|\le M ,$$

and so $|f(x)|\le M$.