Proof of Extreme Value Theorem on a Compact Subset of a Metric Space

Let $\mathcal{T}_{K}$ be the subspace topology on $K$ induced by $\mathcal{T}_{d}$. By Compact Topological Space and Compact Subset §compact-subset, the hypothesis on $K$ says that $(K,\mathcal{T}_{K})$ is compact. The order of $\mathbb{R}$ is total, and the rules of Rules of Arithmetic and Order in an Ordered Field apply to it. Steps 1 and 2 use of $f$ only that it is a map from $K$ to $\mathbb{R}$ that is continuous on $K$ as a map from $(X,d)$ to $(\mathbb{R},d_{\mathbb{R}})$.

Step 1: for every real $c$, the set $V_{c}=\{y\in K:f(y)<c\}$ belongs to $\mathcal{T}_{K}$. Let $\mathcal{S}_{c}$ be the set of all $W\in\mathcal{T}_{d}$ with $K\cap W\subseteq V_{c}$, and put $W_{c}=\bigcup\mathcal{S}_{c}$. As $\mathcal{T}_{d}$ is a topology on $X$ and $\mathcal{S}_{c}\subseteq\mathcal{T}_{d}$, Topological Space §unions gives $W_{c}\in\mathcal{T}_{d}$. If $y\in K\cap W_{c}$, then $y\in W$ for some $W\in\mathcal{S}_{c}$, so $y\in K\cap W\subseteq V_{c}$; hence $K\cap W_{c}\subseteq V_{c}$. Conversely, let $x\in V_{c}$. Then $f(x)<c$, so $\varepsilon=c-f(x)$ satisfies $0<\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. As $f$ is continuous at $x$ relative to $K$, by Continuous Map Between Metric Spaces §continuous, there is a real $\delta>0$ such that every $y\in K$ with $d(x,y)<\delta$ satisfies $d_{\mathbb{R}}(f(y),f(x))<\varepsilon$, that is $|f(y)-f(x)|<c-f(x)$ by The Distance on the Real Numbers §distance. For such $y$, Rules of Arithmetic and Order in an Ordered Field §absolute-value gives $f(y)-f(x)\le|f(y)-f(x)|$, hence $f(y)-f(x)<c-f(x)$, and adding $f(x)$ gives $f(y)<c$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. By Open Ball in a Metric Space §membership this says $K\cap B_{d}(x,\delta)\subseteq V_{c}$, where $B_{d}(x,\delta)$ is the open ball. That ball is open in $(X,d)$ by Open Ball in a Metric Space is Open §open, so it lies in $\mathcal{T}_{d}$ and hence in $\mathcal{S}_{c}$; and $x\in B_{d}(x,\delta)$ because $d(x,x)=0<\delta$ by Metric Space §zero. Thus $x\in K\cap W_{c}$. Hence $V_{c}=K\cap W_{c}$ with $W_{c}\in\mathcal{T}_{d}$ and $V_{c}\in\mathcal{P}(K)$, so $V_{c}\in\mathcal{T}_{K}$ by Subspace Topology §subspace-topology.

Step 2: $f$ attains a maximum. Suppose that there is no $x_{\max}\in K$ with $f(x)\le f(x_{\max})$ for every $x\in K$. Then for every $x\in K$ there is $y\in K$ for which $f(y)\le f(x)$ fails, so that $f(x)<f(y)$, the order being total. Let $\mathcal{U}$ be the set of all $V\in\mathcal{T}_{K}$ with $V=V_{f(y)}$ for some $y\in K$. If $x\in K$ and $y\in K$ satisfies $f(x)<f(y)$, then $x\in V_{f(y)}$, and $V_{f(y)}\in\mathcal{U}$ by Step 1; so $\mathcal{U}$ is a subset of $\mathcal{T}_{K}$ with $K\subseteq\bigcup\mathcal{U}$. As $(K,\mathcal{T}_{K})$ is compact, Compact Topological Space and Compact Subset §compact gives a finite subset $\mathcal{V}$ of $\mathcal{U}$ with $K\subseteq\bigcup\mathcal{V}$. For $V\in\mathcal{V}$ the set $C_{V}=\{y\in K:V=V_{f(y)}\}$ is nonempty, so 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 §choice, applied to the finite set $\mathcal{V}$ and the family $(C_{V})_{V\in\mathcal{V}}$, gives a map $\varphi$ on $\mathcal{V}$ with $\varphi(V)\in C_{V}$ for every $V\in\mathcal{V}$. Let $E$ be the image of $\mathcal{V}$ under the map $V\mapsto f(\varphi(V))$ from $\mathcal{V}$ to $\mathbb{R}$; it is finite 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 §image. Since $K$ is nonempty and $K\subseteq\bigcup\mathcal{V}$, the set $\mathcal{V}$ is nonempty, and so is $E$. 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 §extremes, $E$ has a greatest element $M=\max E$, unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least. Then $M=f(\varphi(V_{0}))$ for some $V_{0}\in\mathcal{V}$; put $y^{*}=\varphi(V_{0})\in K$. As $y^{*}\in K\subseteq\bigcup\mathcal{V}$, there is $V\in\mathcal{V}$ with $y^{*}\in V=V_{f(\varphi(V))}$, so $f(y^{*})<f(\varphi(V))\le M=f(y^{*})$, whence $f(y^{*})<f(y^{*})$, which is impossible. Hence there exists $x_{\max}\in K$ with $f(x)\le f(x_{\max})$ for every $x\in K$.

Step 3: $f$ attains a minimum. Let $g:K\to\mathbb{R}$ be the map $x\mapsto-f(x)$. For $x,y\in K$, Rules of Arithmetic and Order in an Ordered Field §signs gives $g(y)-g(x)=-(f(y)-f(x))$, so $d_{\mathbb{R}}(g(y),g(x))=|f(y)-f(x)|=d_{\mathbb{R}}(f(y),f(x))$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value and The Distance on the Real Numbers §distance. Hence, for every $x\in K$ and every real $\varepsilon>0$, the $\delta$ that Continuous Map Between Metric Spaces §continuous-at provides for $f$ serves for $g$, and $g$ is continuous on $K$ as a map from $(X,d)$ to $(\mathbb{R},d_{\mathbb{R}})$. Steps 1 and 2 applied to $g$ give $x_{\min}\in K$ with $g(x)\le g(x_{\min})$, that is $-f(x)\le-f(x_{\min})$, for every $x\in K$, and then $f(x_{\min})\le f(x)$ by Rules of Arithmetic and Order in an Ordered Field §order-negative.

Steps 2 and 3 together give the assertion.