Reminder: An isometry of metric spaces $(X,d)$, $(Y,d')$ is a function
$$f:X \to Y$$ Such that for any $s,t \in X$ we have
$$d(s,t) = d(f(s),f(t))$$
i.e. It preserves all distances.
A homeomorphism of metric spaces $(X,d)$, $(Y,d')$ is a bijection from $X$ to $Y$ which is continuous and has continuous inverse.
Although every surjective isometry is a homeomorphism, it is not true that every homeomorphism is an isometry, however if we are given a metric space $(X,d)$ and two subpaces $(S,d)$ and $(K,d)$ of $X$, and we know that for any other subset $J \subseteq X$, $J\cap S$ is homeomorphic to $J\cap K$, then there is an isomemtry $S\to K$ (In fact one can simply conclude that $S = K$) However, we can try to strengthen this theorem and assume that for every open set $J\subseteq X$, there is a homeomorphism $J\cap S \to J\cap K$.
Question: Is this still sufficient to conclude that there is an isometry between $S$ and $J$ ?
No. An example is $X={\mathbb R}$ with the standard metric, while $S={\mathbb Q}$, $K=S\setminus \{0\}$.