Proof of Extreme Value Theorem on a Closed Real Interval

Since $a\le a\le b$, the point $a$ belongs to the closed interval $[a,b]$, so $[a,b]$ is nonempty. By Closed Interval $[a,b]$ is Compact in $\mathbb{R}$, $[a,b]$ is compact in $\mathbb{R}$ equipped with the topology consisting of the subsets that are open in $(\mathbb{R},d_{\mathbb{R}})$, which is a topology by Metric Open Sets Form a Topology.

We verify the continuity property assumed in Extreme Value Theorem on a Compact Subset of a Metric Space for the metric space $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Line, the nonempty compact subset $K=[a,b]$, and the function $f$. Let $x\in[a,b]$ and let $\varepsilon>0$ be real. Since $f$ is continuous at $x$ relative to $[a,b]$, there exists a real $\delta>0$ such that every $y\in[a,b]$ with $d_{\mathbb{R}}(x,y)<\delta$ satisfies $d_{\mathbb{R}}(f(y),f(x))<\varepsilon$. Since $d_{\mathbb{R}}(s,t)=|s-t|$ for all real $s,t$ by The Absolute Value Metric on the Real Line, where $|\cdot|$ is the absolute value on $\mathbb{R}$, this says precisely: every $y\in[a,b]$ with $d_{\mathbb{R}}(x,y)<\delta$ satisfies $|f(y)-f(x)|<\varepsilon$. This is exactly the continuity property required of $f$ in Extreme Value Theorem on a Compact Subset of a Metric Space.

By Extreme Value Theorem on a Compact Subset of a Metric Space there exist $x_{\min},x_{\max}\in[a,b]$ such that

$$f(x_{\min})\le f(x)\le f(x_{\max})\qquad\text{for every }x\in[a,b] . \qquad\blacksquare$$