Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a smooth map such that $|Df(p)(u)| = |u|$ for all $p, u \in \mathbb{R}^n$. Prove that $f$ preserves distances, i.e. $|f(p) - f(q)| = |p - q|$ for all $p,q \in \mathbb{R}^n$.
I've already shown that $Df(p)$ is an isomorphism and a linear isometry, and therefore $f$ is a diffeomorphism. How do I go from here to showing that $f$ preserves distances?
Let $p$, $q\in{\mathbb R}^n$ be given, let $$\gamma: \quad t\mapsto x(t):=(1-t)p+t q\qquad(0\leq t\leq1)$$ be the segment connecting $p$ with $q$, and let $$\gamma':\quad t\mapsto g(t):=f\bigl(x(t)\bigr)\qquad(0\leq t\leq 1)$$ be its image curve. Then $g'(t)=df\bigl(x(t)\bigr).x'(t)$ and therefore $|g'(t)|=|x'(t)|$, by assumption on $df$. It follows that $$|f(p)-f(q)|\leq L(\gamma')=L(\gamma)=|p-q|\ ,$$ whatever $p$, $q\in{\mathbb R}^n$.
On the other hand, our $f$ is a local diffeomorphism, hence each point $p$ has a neighborhood $U$ which is bijectively mapped onto a neighborhood $V$ of $p'=f(p)$. Points $q'$ nearby $p'$ can be joined to $p'$ with a segment that has a preimage in $U$, so that (applying the above argument to the local $f^{-1}$) we now also have $|p-q|=|f^{-1}(p')-f^{-1}(q')|\leq |p'-q'|$. This shows that $f$ is a local isometry, and allows to conclude that the image of a line is straight all the way. That should suffice.