There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold $\newcommand{\al}{\alpha}$
$\newcommand{\ga}{\gamma}$
$\newcommand{\e}{\epsilon}$
Let $X,Y$ be Riemannian manifolds, such that $\dim(X) > \dim(Y)$.
I am trying to prove the following statement (mentioned by Gromov in his book on metric geometry):
There is no arcwise isometry (i.e length preserving map) from $X$ to $Y$.
However, the naive attempt to prove this hits an obstacle which I do not see how to pass:
Suppose by contradiction $f:X \to Y$ is an arcwise isometry. Then $f$ is $1$-Lipschitz, hence differeniable almost everywhere (by Rademacher's theorem).
Question: Let $p \in X$ be a point where $f$ is differentiable. Is $df_p:T_pX \to T_{f(p)}Y$ an isometry?
(This will imply the claim of course).
Here is what happens when trying to show this naively:
Let $v \in T_pX$, and let $\al:[0,1] \to X$ be a smooth path s.t $\al(0)=p,\dot \al(0)=v$.
Then $|\dot \alpha(s)| = |\dot \alpha(0)|+\Delta(s)=|v|+ \Delta(s)$ where $\lim_{s \to 0} \Delta(s) =\Delta(0)= 0$, thus
$$ (1) \, \,  L(\alpha|_{[0,t]})=\int_0^t |\dot \alpha(s)| ds=t|v|+\int_0^t  \Delta(s) ds.$$
$\al$ is Lipschitz and $f$ is $1$-Lipschitz, so $\ga:= f \circ \al$ is Lipschitz. By theorem 2.7.6 in the book ``A course in metric geometry'' ( Burago,Burago,Ivanov) it follows that:
$$ (2) \, \, L(\ga|_{[0,t]})=\int_0^t \nu_{\ga}(s) ds, $$
where $\nu_{\ga}(s):=\lim_{\e \to 0} \frac{d\left( \ga(s),\ga(s+\e) \right)}{|\e|}$ is the speed of $\ga$.
Note that $\nu_{\ga}(0)=  |\dot \ga(0)|=|df_p(v)|$.
Using the assumption $f$ preserves lengths, we would now like to compare $(1),(2)$ and take derivatives at $t=0$, to get $$|v|=\frac{d}{dt}\left. \right|_{t=0} L(\al|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]})=|df_p(v)|.$$
However, it seems that the last equality is false in general; Even when the speed of a Lipschitz curve and the derivative of its length exist at a point, they do not need to be equal. 
It seems that the only thing we can say is that $ \frac{d}{dt}\left. \right|_{t=0} L(\ga|_{[0,t]}) \ge \nu_{\ga}(0) =|df_p(v)|$, so we are left with $|v| \ge |df_p(v)|$ which doesn't help.
Is there a way to "fix" the proof above?
 A: Your proof is correct, but you need to add words "amost everywhere" at ane more place.
We use Rademacher's theorem and lemma about length of curve, which says that if a curve parametrized by length then its velocity is 1 almost everywhere, see 2.7.4 in Metric Geometry by Burgo, Burago and Ivanov.
Fix a chart $U_{\subset\mathbb{R}^n}\to X$ and a vector $v\in \mathbb{R}^n$.
Let $g$ be the induced metric on $U$.
Note that from the lemma we get that $|d_pv|=|v|_g$ almost everywhere.
Repeat the same for $N=\tfrac12\cdot n\cdot(n+1)$ vectors $v_1,\dots,v_N$ in general position.
We get $|d_pv_i|=|v_i|_g$ for almost all $p$ and all $i$.
It follows that for all $w$,the identity $|d_pw|=|w|_g$ holds for almost all $p$.
Hence $\dim X \le  \dim Y$.
You may want to check 


*

*Problem "Length-preserving map"  in may collection of problems

*My paper which discuss in particular length-preserving maps and dimension.

A: I am completenig some details based on Anton's answer:
We prove the following theorem:
Let $X,Y$ be Riemannian manifolds, and let $f:X \to Y$ be a length preserving map. Then $df$ is an isometry almost everywhere.

The proof is composed of 3 steps:
Step I:
It is enough to prove the claim where $X=(\mathbb{R}^n,g)$ where $g$ is an arbitrary metric.
Step II:
Assume step I, i.e we consider $f:(\mathbb{R}^n,g) \to Y$.
Let $v \in \mathbb{R}^n$ fixed . Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.
Step III:
Conclude that $df$ is an isometry almost everywhere.

We begin with proofs of steps I,III, since these are easy.
Proof of step I:
$f$ is $1$-Lipschitz, hence by Rademacher's Thm it is differentiable a.e.
Since any manifold admits a countable cover of charts, we finished.
Proof of step III:
$f:(\mathbb{R}^n,g) \to Y$.
Let $v_1,...,v_n$ be a basis for $\mathbb{R}^n$. By applying step II for every $v \in Q=\{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, we get that 
$|df_x(v)|_{f(x)}=|v|_x$ for all $v \in Q$ and for almost every $x \in \mathbb{R}^n$. Now it is a simple linear algebra fact that at all these points $x$, $df_x$ is an isometry. Indeed,
$$ \langle df_x(v_i),df_x(v_j) \rangle = \frac{1}{2}(|df_x(v_i)+df_x(v_j)|^2 - |df_x(v_i)|^2 - |df_x(v_j)|^2) = \frac{1}{2}(|df_x(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)  $$
$$ = \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle.$$
Comment: Here we see why restricting to a chart was so helpful! We can use the "same" basis vectors $v_i$ for different points $x$ (on a general non-parallelizable manifold, this is meaningless).

Proof of step II:
Fix $v \in \mathbb{R}^n$. Then $|df_x(v)|_{f(x)}=|v|_x$ for almost every $x \in \mathbb{R}^n$.
Proof:
Fix $x \in \mathbb{R}^n$, and define $\alpha_x(t)=x+tv$.
Putting $\gamma_x=f \circ \alpha_x$ (this is a Lipschitz path), we get by theorem 2.7.6 that $$ (1): \, \, \nu_{\alpha_x}(t) =\frac{d}{dt}\left. \right|_{t=0} L(\alpha_x|_{[0,t]})=\frac{d}{dt}\left. \right|_{t=0} L(\ga_x|_{[0,t]})=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$
For any path $\beta$ in a Riemannian manifold, at every point of differentiability, it holds that $\nu_{\beta}(t)=|\dot \beta(t)|_{\beta(t)}$.
Specializing this for $\beta = \alpha_x$ (and combining $(1)$) we obtain
$$ (1)': \, \, |v|_{\alpha_x(t)}=\nu_{\ga_x}(t) \, \, \text{ for almost every $t$}$$
Specializing this for $\beta = \gamma_x$ leads to 
observation I:
$\nu_{\gamma_x}(t)=|df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}$  for those $t$ where $f$ is differentiable at $\alpha_x(t)$.

We define the sets
$ B:=\{(x,t) \in \mathbb{R}^n \times \mathbb{R} | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \,  \text{ is differentiable at } \, \alpha_x(t) \}  \subseteq \mathbb{R}^{n+1}, $
$B_x=\{ t \in \mathbb{R} |  \, \, (x,t) \in B \} \subseteq \mathbb{R}$ and
$B^t= \{x\in\Bbb R^n: (x,t)\in B\} \subseteq \mathbb{R}^n$. 
We also define $h:\mathbb{R}^{n+1} \to \mathbb{R}^n$ by $h(x,t)=\alpha_x(t)$.
$h(B)=\{\alpha_x(t) | \, \, \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} \, \, \text{ and } \, f \,  \text{ is differentiable at } \, \alpha_x(t) \}=\{\alpha_x(t) | \, \, |df_{\alpha_x(t)}(v)|_{f(\alpha_x(t))}=|v|_{\alpha_x(t)}  \}$ 
where the last equality follows from observation I.
Hence, it is enough to show $h(B)^c$ has measure zero in $\mathbb{R}^n$.
This essentially follows from Tonelli's theorem:
Since $h$ is surjective, $h(B)^c \subseteq h(B^c)$. Thus, it is enough to show $\mu(B^c)=0$ in $\mathbb{R}^{n+1}$.
$$ B^c=\{  (x,t)  | \, \, \neg ( \nu_{\gamma_x}(t)=|v|_{\alpha_x(t)} ) \} \cup \{ (x,t)  | \, \,  f \, \text{ is not differentiable at } \, \alpha_x(t) \}:=A \cup D$$
$(1)'$ implies that $\lambda_1(A_x)=0$ for every $x \in \mathbb{R}^n$, hence (by Tonelli) $\mu(A)=0$.
$D^t = \{ x \in \mathbb{R}^n | f \, \text{ is not differentiable at } \, x+tv \} = W -tv$ (where $W \subseteq \mathbb{R}^n$ is the set of points of non-differentiability of $f$), so $\lambda_n(D^t)=\lambda_n(W)=0$
Again, Tonelli implies $\mu(D)=0$.
