I think it's a simple proof, but I've got no idea how to start.
Let $\phi :M\rightarrow N $ be an isometry between two surfaces. Now show that the length of the curve does not change by mapping it with $\phi$. Vice-versa a diffeomorphism $\phi :M\rightarrow N $ that does not change the length of the curve by mapping with $\phi$ is an isometry.
Let $g$ be the metric on $M$, $h$ be the metric on $N$ and $\gamma\colon [0,1]\rightarrow M$ be a smooth curve on $M$, then one has: $$\begin{align}\ell(\phi\circ\gamma)&:=\int_0^1\sqrt{h_{\phi\circ\gamma(t)}((\phi\circ\gamma)'(t),(\phi\circ\gamma)'(t))}\,\mathrm{d}t,\\&=\int_0^1\sqrt{h_{\phi\circ\gamma(t)}(T_{\gamma(t)}\phi(\gamma'(t)),T_{\gamma(t)}\phi(\gamma'(t)))}\,\mathrm{d}t,\tag{1}\\&=\int_0^1\sqrt{(\phi^*h)_{\gamma(t)}(\gamma'(t),\gamma'(t))}\,\mathrm{d}t,&\\&=\int_0^1\sqrt{g_{\gamma(t)}(\gamma'(t),\gamma'(t))}\tag{2},\\&=:\ell(\gamma),\end{align}$$ whence the result.
The equality $(1)$ follows from the chain rule and $(2)$ is the definition of $\phi$ being an isometry.