I need proof verification on the following theorem (Lee intro to Riemannian Manifolds Prop 2.51):
Let $\varphi : (M,g) \longrightarrow (\tilde{M},\tilde{g})$ be an isometry between connected riemannian manifolds, then for all $x,y \in M$ we have
$$d_g(x,y)=d_{\tilde{g}}(\varphi(x),\varphi(y)),$$
$d_g$ and $d_{\tilde{g}}$ being the Riemann distance function on $M$ and $\tilde{M}$ respectively.
I start by proving that if $\varphi$ is a local isometry, then
$$d_g(x,y)\geq d_{\tilde{g}}(\varphi(x),\varphi(y)) \quad \forall x,y\in M.$$
If then $\varphi$ is an isometry, then both $\varphi$ and its inverse are local isometry and I can conclude that equality holds.
Let $x,y \in M$ and $\gamma : [a,b] \longrightarrow M$ be an admissible curve (piecewise regular) joining $x$ and $y$ in $M$, and here I'm assuming $\gamma$ is injective, and let $\Gamma = \gamma([a,b])$. Observe that $\Gamma$ is compact by continuity of $\gamma$.
Since $\varphi$ is a local isometry, for each $p\in \Gamma$ we find a neighborhood $U_p$ in $M$ such that $\varphi\big|_{U_p}$ is an isometry. Since $\{U_p\}_{p\in \Gamma}$ is an open cover of $\Gamma$ we find a finite subcover $\{U_0,\ldots,U_n\}$.
Define $\Gamma_i = U_i \cap \Gamma$. What I have in mind here is that, after suitable reordering of the indexes and refinement, that $\Gamma_i$ is a partition of $\Gamma$, meaning $[t_i,t_{i+1}]=\gamma^{-1}(\overline{\Gamma_i})$ and $a=t_0<\ldots<t_{n+1}=b$ is a partition of $[a,b]$. I can now write
\begin{align}
L_g(\gamma)&=\sum_{i=0}^{n} L_g(\gamma|_{[t_i,t_{i+1}]})=\sum_{i=0}^{n} L_{\tilde{g}}((\varphi \circ \gamma)|_{[t_i,t_{i+1}]}) \tag{1}\\
&\geq \sum_{i=0}^{n} d_{\tilde{g}}(\varphi(x_i),\varphi(x_{i+1})) \geq d_{\tilde{g}}(\varphi(x),\varphi(y)) \tag{2}
\end{align}
where $x_i=\gamma(t_i), \quad i=1,\ldots,n$. The first equality follows by definition of local isometry and the second inequality from the triangle inequality.
Since $\gamma$ is arbitrary I conclude that $d_g(x,y)\geq d_{\tilde{g}}(\varphi(x),\varphi(y)).$
In the exercise Lee asks for a counterexample showing that local isometries don't preserve distances, I thought of
$$\varphi : [0,2\pi] \longrightarrow \mathbb{S}^1, \quad \varphi(t)=(\cos t,\sin t),$$
which is clearly a local isometry but not injective.
Best Answer
Your proof works, but there is a much shorter way. Let $\varphi\colon (M,g) \to (N,h)$ be an isometry. Fix $x$ and $y$ in $M$. Let $\Gamma_{x,y}$ be the set of piecewise $\mathcal{C}^1$ curves joining $x$ and $y$ (say, parametrized on $[0,1]$). For $\varphi$ is a diffeomorphism, the map $ F\colon \gamma \in \Gamma_{x,y} \longmapsto \varphi \circ \gamma \in \Gamma_{\varphi(x),\varphi(y)} $ is a bijection. Moreover, since $\varphi$ is an isometry, for $\varphi\in \Gamma_{x,y}$, one has $ \ell(\gamma) = \int_0^1\|\gamma'\|_g = \int_0^1 \|(\varphi\circ\gamma)'\|_h = \ell(\varphi\circ \gamma). $ It follows that $$ d_g(x,y) = \inf_{\gamma \in \Gamma_{x,y}}\ell(\gamma) = \inf_{\gamma\in \Gamma_{x,y}}\ell(\varphi\circ \gamma) = \inf_{\sigma \in \Gamma_{\varphi(x),\varphi(y)}} \ell(\sigma) = d_h(\varphi(x),\varphi(y)). $$
Also, your counterexample works perfectly (if you want to use manifolds without boundary, just extend it to $\Bbb R$).