Given a compact metric space $(X,d)$, we consider $Iso(X,d)$ with metric $\rho$ such that
$\lim _{n \rightarrow \infty} \rho(h_n, h) =0 \iff \forall x \in X: \lim _{n \rightarrow \infty} d(h_n(x), h(x))=0$.
Could you tell me how to prove that if we have a $\rho$-Cauchy sequence of functions $h_n \in Y$, then for each $x\in X$, we have that $h_n(x)$ is a $d$-Cauchy sequence of points of $X$?
I'm just starting studying topology, so I would appreciate all your help.
It seems that I have a following sketch of the proof. By Exercise 9.4 from [Kech], $\operatorname{Iso}(X)$ is
compact in the topology of uniform convergence and, hence in a weaker topology of pointwise convergence too. By Theorem 4.3.28 from [Eng], every metric on a compact space is complete. Therefore a $\rho$-Cauchy sequence $\{h_n\}$ converges. Then by the property of the metric $\rho$ from the second row of the question, for each $x\in X$, we have that a sequence $\{h_n(x)\}$ is $d$-Cauchy (even, $d$-convergent) sequence of points of $X$.
Of course, you can try to prove the referenced results directly. For instance, to prove that $\operatorname{Iso}(X)$ is a compact in the topology of pointwise convergence it suffices to show that $\operatorname{Iso}(X)$ is a closed subset of the Tychonoff product $X^X$. To prove that every Cauchy sequence $\{h_n\}$ in a metric compact $Y$ is convergent it suffice to show that $\{h_n\}$ has a cluster point $h$, and, since the space $Y$ is first countable, there is a convergent to $h$ subsequnce $\{h_{n_k}\}$ of the sequence $\{h_n\}$. Since the sequence $\{h_n\}$ is Cauchy, it converges to $h$ too. Moreover, atfter writing the answer I noticed in Related a question "Compact metric space group $Iso(X,d)$ is also compact", and I tried to clarify some details in my answer there.
References
[Eng] Ryszard Engelking. General Topology (Russian version, 1986).
[Kech] Alexander S. Kechris. Classical Descriptive Set Theory , Springer, 1995.