Proof of The Continuous Image of a Compact Subset of a Metric Space is Compact

Only a finite choice is used: the cover of $K$ below is defined by a formula, and the passage back to $f(K)$ selects from finitely many nonempty sets, 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 (xi). Throughout, compactness of a subset means compactness of the subset with its subspace topology, as in Compact Topological Space and Compact Subset (ii), and for $U\subseteq Y$ we write

$$f^{-1}(U)=\{x\in K:f(x)\in U\}.$$

The heart of the proof is that preimages of relatively open sets are relatively open. Let $U$ be open in $f(K)$; by Subspace Topology (ii), $U=f(K)\cap V$ for some $V\in\mathcal{T}_{\rho}$. Let $O$ be the union of all open balls $B_{d}(x,r)$ with $x\in K$, $r>0$ and $B_{d}(x,r)\cap K\subseteq f^{-1}(U)$. Open balls are open, so $O\in\mathcal{T}_{d}$ by Topological Space (iii), and $O\cap K\subseteq f^{-1}(U)$ by construction. For the reverse inclusion, let $x\in f^{-1}(U)$. First, since $V$ is open and contains $f(x)$, choose $\varepsilon>0$ with $B_{\rho}(f(x),\varepsilon)\subseteq V$ (Open Subset of a Metric Space (i)); then, for this $\varepsilon$, continuity of $f$ at $x$ (Continuous Map Between Metric Spaces (i)) gives $\delta>0$ such that

$$\rho(f(y),f(x))<\varepsilon\qquad\text{for every }y\in K\text{ with }d(x,y)<\delta.$$

For such $y$ we get $f(y)\in f(K)\cap V=U$; so $B_{d}(x,\delta)\cap K\subseteq f^{-1}(U)$ and $x\in O$. Therefore

$$f^{-1}(U)=K\cap O,$$

which is open in $K$ by Subspace Topology (ii).

Now let $\mathcal{U}$ be a set of open subsets of $f(K)$ covering $f(K)$, and let $\mathcal{W}$ be the set of all $W\in\mathcal{P}(K)$ of the form $W=f^{-1}(U)$ with $U\in\mathcal{U}$. By what was just shown, $\mathcal{W}$ consists of open subsets of $K$, and it covers $K$: for $x\in K$ the point $f(x)$ lies in some $U\in\mathcal{U}$, and then $x\in f^{-1}(U)$. Since $K$ is compact, some finite $\mathcal{V}\subseteq\mathcal{W}$ covers $K$ (Compact Topological Space and Compact Subset (i)).

It remains to return to $f(K)$. For each $W\in\mathcal{V}$ the set $\{U\in\mathcal{U}:f^{-1}(U)=W\}$ 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 (xi) gives a function $g$ on $\mathcal{V}$ with $g(W)\in\mathcal{U}$ and $f^{-1}(g(W))=W$ for every $W\in\mathcal{V}$. Its image $\mathcal{U}'=\{g(W):W\in\mathcal{V}\}$ is a finite subset of $\mathcal{U}$ (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 (vi)), and it covers $f(K)$: a point of $f(K)$ is $f(x)$ with $x\in K$, this $x$ lies in some $W\in\mathcal{V}$, and then $f(x)\in g(W)$. Hence $f(K)$ is compact in $(Y,\mathcal{T}_{\rho})$ by Compact Topological Space and Compact Subset (i) and Compact Topological Space and Compact Subset (ii), which proves (i).