Isomorphism isometries between finite subsets , implies isomorphism isometry between compact metric spaces 
Let's $(X_1,d_1), (X_2,d_2)$ be compact metric spaces such that for every finite subset of $X_1$ like $A$ (respectively any finite subset of $X_2$ like $B$ ) there exists a finite subset of $X_2$ like $B$ ( respectively exists a finite subset of $X_1$ like $A$ ) such that there exists a isomorphism isometry between $A$ and $B$.

Show that there exists an isomorphism isometry between $X_1$ and $X_2$
 A: Every compact metric space is separable, so let $D=\{x_k:k\in\Bbb N\}$ be a dense subset of $X_1$. For each $n\in\Bbb N$ let $D_n=\{x_k\in D:k\le n\}$; $D_n$ is finite, so there is an isometry $f_n:D_n\to X_2$. For each $k,n\in\Bbb N$ with $k\le n$ let $y_k^n=f_n(x_k)$, so that $f_n[D_n]=\{y_k^n:k\le n\}$. Note that for any $k,\ell,n\in\Bbb N$ with $k,\ell\le n$ we have $d_2(y_k^n,y_\ell^n)=d_1(x_k,x_\ell)$.
Let $N_0=\Bbb N$. Every compact metric space is sequentially compact, so there is an infinite $N_1\subseteq N_0$ such that the subsequence $\langle y_0^n:n\in N_1\rangle$ of $\langle y_0^n:n\in N_0\rangle$ converges to some $y_0\in X_2$. Similarly, there is an infinite $N_2\subseteq N_1$ such that the subsequence $\langle y_1^n:n\in N_2\rangle$ of $\langle y_1^n:n\in N_1\rangle$ converges to some $y_1\in X_2$. In general, given an infinite $N_k\subseteq\Bbb N$, there is an infinite $N_{k+1}\subseteq N_k$ such that the subsequence $\langle x_k^n:n\in N_{k+1}\rangle$ of $\langle y_k^n:n\in N_k$ converges to some $y_k\in X_2$.
Now define $f:D\to X_2$ by $f(x_k)=y_k$ for each $k\in\Bbb N$. Fix $k,\ell\in\Bbb N$, and let $m=1+\max\{k,\ell\}$. Then $\langle y_k^n:n\in N_m\rangle\to y_k$, $\langle y_\ell^n:n\in N_m\rangle\to y_\ell$, and $d_2(y_k^n,y_\ell^n)=d_1(x_k,x_\ell)$ for each $n\in N_m$, so $d_2(y_k,y_\ell)=d_1(x_k,x_\ell)$. Thus, $f$ is an isometry of $D$ onto $f[D]=\{y_k:k\in\Bbb N\}$.
Let $p\in X_1$. There is a sequence $\sigma=\langle p_k:k\in\Bbb N\rangle$ in $D$ converging to $p$. Now $\sigma$ is a Cauchy sequence, and $f$ is an isometry, so $\langle f(p_k):k\in\Bbb N\rangle$ is a Cauchy sequence in $X_2$ and therefore convergent to some $q\in X_2$. I leave it to you to show that $q$ is independent of the choice of sequence $\sigma$ in $D$ converging to $p$, so that if we set $F(p)=q$, the function $F:X_1\to X_2$ is well-defined. I also leave to you the straightforward verification that $F$ is an isometry from $X_1$ to $F[X_1]$; the only problem is that $f[X_1]$ might not be all of $X_2$.
However, the same argument shows that there is an isometric embedding $G$ of $X_2$ into $X_1$. Let $h=G\circ F:X_1\to X_1$; clearly $h$ is another isometry. If $F$ is not surjective, neither is $h$, and there is a point $p\in X_1\setminus h[X_1]$. The map $h$ is continuous, and $X_1$ is compact, so $h[X_1]$ is compact and therefore closed, and there is an $r>0$ such that $d_1(p,x)\ge r$ for all $x\in h[X_1]$. Let $p_0=p$, and for each $k\in\Bbb N$ let $p_{k+1}=h(p_k)$. Fix $n\in\Bbb Z^+$ arbitrarily. Clearly $p_n\in h[X_1]$, so $d_1(p_0,p_n)\ge r$. For each $k\in\Bbb N$ apply the isometry $$h^k=\underbrace{h\circ\ldots\circ h}_k$$ to $p_0$ and $p_n$ to find that $d_1(p_k,p_{n+k})=d_1(p_0,p_n)\ge r$. Thus, $d_1(p_k,p_\ell)\ge r$ whenever $k\ne\ell$, and the sequence $\langle p_k:k\in\Bbb N\rangle$ has no Cauchy subsequence and hence no convergent subsequence. $X_1$ is sequentially compact, so this is impossible, and $F$ must in fact be surjective, as desired.
