Proof of A Distance-Preserving Bijection is a Homeomorphism

Claim 1. Let $x\in X$ and let $r>0$ be a real number. If $v\in B_{d_{X}}(x,r)$, that is $d_{X}(x,v)<r$ by the definition of the open ball, then $d_{Y}(\Phi(x),\Phi(v))=d_{X}(x,v)<r$, so $\Phi(v)\in B_{d_{Y}}(\Phi(x),r)$. This proves the inclusion from left to right. Conversely, let $y\in B_{d_{Y}}(\Phi(x),r)$. Since $\Phi$ is a bijection there is $v\in X$ with $y=\Phi(v)$, and then

so $v\in B_{d_{X}}(x,r)$ and hence $y\in\Phi\bigl(B_{d_{X}}(x,r)\bigr)$.

Claim 2. We first show that $\Phi$ is continuous. Let $W\in\mathcal{T}_{d_{Y}}$ and put $U=\{x\in X:\Phi(x)\in W\}$. Let $x\in U$. Since $W$ is open in $(Y,d_{Y})$ and $\Phi(x)\in W$, there is a real $r>0$ with $B_{d_{Y}}(\Phi(x),r)\subseteq W$. By claim 1, $\Phi\bigl(B_{d_{X}}(x,r)\bigr)=B_{d_{Y}}(\Phi(x),r)\subseteq W$, so every $v\in B_{d_{X}}(x,r)$ satisfies $\Phi(v)\in W$, that is $B_{d_{X}}(x,r)\subseteq U$. Since $x\in U$ was arbitrary, $U$ is open in $(X,d_{X})$, that is $U\in\mathcal{T}_{d_{X}}$. Hence $\Phi$ is continuous.

The inverse map $\Phi^{-1}:Y\to X$ is again a bijection, and it preserves distances: for $y,y'\in Y$, applying the hypothesis to $u=\Phi^{-1}(y)$ and $v=\Phi^{-1}(y')$ gives

So the argument just given, applied to $\Phi^{-1}$ in place of $\Phi$ and with the roles of $(X,d_{X})$ and $(Y,d_{Y})$ interchanged, shows that $\Phi^{-1}$ is continuous.

Claim 3. Suppose first that $A$ is compact in $X$. By claim 2 the map $\Phi$ is continuous, so claim 2 of Restriction of a Continuous Map, and Continuous Images of Compact Subsets shows that $\Phi(A)$ is compact in $Y$. Suppose conversely that $\Phi(A)$ is compact in $Y$. By claim 2 the map $\Phi^{-1}$ is continuous, so the same result, applied to $\Phi^{-1}$ and the subset $\Phi(A)$ of $Y$, shows that $\{\Phi^{-1}(y):y\in\Phi(A)\}$ is compact in $X$. Since $\Phi$ is a bijection, that set is $A$.