I stumbled across the previous question (I rewrote it here in case the original one isn't edited yet), but it was closed. I found it interesting, so I'll give the little reasoning that I did.
If we consider $\varphi:S\to\tilde S$ a difeomorphism, we know that $\varphi$ is an isometry if and only if it preserves lengths by arc of the parameter curves in $S$ $(\star)$. Since the intrinsic distance is defined as follows:
\begin{equation*} d(p,q) = \inf\{l(\alpha_{p,q})\; |\; \alpha_{p,q}\in D^\infty(p,q)\},\;\;\forall \; p,q\in S, \end{equation*} where $D^\infty(p,q)$ is the set of parameter curves that are differentiable by pieces and join $p$ to $q$.
Pardon my abuse of notation, but from $(\star)$ we know that $l(\alpha_{p,q}) = l(\alpha_{\varphi(p),\varphi(q)})$, so in particular $\inf \left\{l(\alpha_{p,q})\right\} = \inf \left\{l(\alpha_{\varphi(p),\varphi(q)})\right\}$, so $d(p,q) = d(\varphi(p),\varphi(q))$.
Is this argument correct?
P.S.: I've seen that in Wikipedia the isometry is defined as a map that preserves distances. Here is the definition that my professor has given us in her lecture:
Definition: Let $S$ and $\tilde S$ be two surfaces. We say that $\varphi:S\to\tilde S$ is a isometry if $\varphi$ is a diffeomorphism and $d\varphi_p:T_pS\to T_p\tilde S$ is a linnear isometry.
As stated by Andrew D. Hwang in the comment, my answer is correct. I will rewrite his answer in the same comment to add it here and mark this question as answered:
Consider $\varphi:S\to\tilde S$ an isometry, and $d:S\times S\to \Bbb R$ the intrinsic distance. We can argue that, for $p,q\in S$, $d(\varphi(p),\varphi(q))\le d(p,q)^{(*)}$ because each path from $p$ to $q$ in $S$ gives rise to a path of the same length from $\varphi(p)$ to $\varphi(q)$.
But if $\varphi$ is an isometry, $\varphi^{-1}:\tilde S\to S$ is also an isometry, so if we consider $\varphi(p),\varphi(q)\in\tilde S$ (i pick these points for convenience sake), we have that $d(\varphi^{-1}(\varphi(p)),\varphi^{-1}(\varphi(q)))\le d(\varphi^{-1}(p),\varphi^{-1}(q))\Rightarrow d(p,q)\le d(\varphi^{-1}(p),\varphi^{-1}(q))$, so if we reverse the roles of $S$ and $\tilde S$, we get that $d(p,q)\le d(\varphi(p),\varphi(q))^{(**)}$.
From $(*)$ and $(**)$ we get what we were looking for.
Thanks again to Andrew D. Hwang for this answer!