Nonexpanding map between manifolds decreases volume? Let $M,N$ be diffeomorphic compact Riemannian manifolds, and let $f:M \to N$ be a nonexpanding map (i.e Lipschitz with constant $1$). Assume that
$(1)$ $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $d(f(p),f(q)) < d(p,q)$. 
$(2)$ The image $f(M)$ is a submanifold of $N$. (Note I do not assume $f$ is smooth).
Is it true that $\operatorname{Vol}(f(M))<\operatorname{Vol}(M)$?
If it helps, we can assume for start $M,N$ have empty boundary.
Note that if we do not assume $M,N$ are diffeomorphic then the answer is negative:
$f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$ is strictly nonexpanding but $\operatorname{Vol}(f([0,2\pi])=\operatorname{Vol}(\mathbb{S}^1)=\operatorname{Vol}([0,2\pi])$.
Partial result:
$(1)$ In the case where $M=N$ (as Riemannian manifolds), the answer is positive.
Assume otherwise; Then $\operatorname{Vol}(f(M))=\operatorname{Vol}(M)=\operatorname{Vol}(N)$, hence $f(M)=N$, i.e $f$ is surjective. (Otherwise $f(M)$ will be a closed subset of $N$, strictly contained in $N$, contradicting the equality of volumes).
So,  $f$ is a surjective nonexpanding map from a compact metric space to itself, thus an isometry. (See Burago-Burago-Ivanov's "A course in metric geometry", theorem 1.6.15).
$(2)$ In the case the manifolds are one-dimensional, and $f$ is surjective, the answer is positive:
Assume otherwise. Then $\operatorname{Vol}(N)=\operatorname{Vol}(M)$. Since every two compact connected one-dimensional Riemannian manifolds with equal volumes are isometric, there exists an isometry $\phi:N \to M$.
Thus, $f \circ \phi:N \to N$ is a surjective nonexpanding map from a compact metric space to itself, hence an isometry. 

Result $(1)$ suggests it might be easier to handle the case where $\operatorname{Vol}(M)=\operatorname{Vol}(N)$. The question then becomes equivalent to the following one:
Can a strictly nonexpanding map between two compact Riemannian manifolds of the same volume be surjective? 
 A: Here is @AntonMalyshev counterexample in detail (for the case of manifolds with corners):
Let $M$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,1/3]$. This is a manifold with corners (for example, a small enough neighborhood of the point $(0,1/3)\sim (1,1/3)$ is diffeomorphic to $[0,\infty)\times[0,\infty)$).
The distance between $(0,1)$ and $(1,1)$ can easily be verified to be $1$.
Let $N$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,2/3]$. Here the distance between $(0,1)$ and $(1,1)$ is at most $2/3$, as shown by the curve
$$
\gamma(t) = 
\begin{cases}
(0,1)(1-t) + (0,2/3)t & t\in[0,1] \\
(1,2/3)(2-t) + (1,1)(t-1) & t\in[1,2].
\end{cases}
$$
Obviously $M$ and $N$ are diffeomorphic as manifolds with corners, and have the same volume.
The trivial map $M\to N$ is surjective, volume preserving and strictly non-expending according to your definition.
A: If you allow boundaries and allow $\operatorname{Vol}(f(M))<\operatorname{Vol}(N),$ the answer is negative. Take $M$ to be a connected non-convex subset in $\mathbb R^2$ with smooth boundary such as an annulus, and embed it in a convex set in a scaled-up version of itself, $N.$ The embedding decreases distances between points that were not connected by straight lines in $M,$ but preserves volume.

If $M$ is complete, then the answer is positive. We don't need to assume it is diffeomorphic to $N.$
Assume $f:M\to N$ has Lipschitz constant $1,$ and $\operatorname{Vol}(f(M))=\operatorname{Vol}(M).$ The aim is to show $f$ is distance preserving. The restriction $f|_{M^o}$ to the interior of $M$ is injective and locally distance preserving by the area formula. The image is closed and open, and any path in $f(M)$ can be pulled back to $M,$ so $f$ is distance-preserving.

Now allow $M$ and $N$ to be compact manifolds with boundary, but assume they have equal volume. The answer is again positive.
As before $f|_{M^o}$ is injective and locally distance preserving, but now it also has dense image. In particular $f$ is surjective. And by flowing along inward pointing geodesics as in https://mathoverflow.net/q/253994 we find that $f$ is smooth on the whole of $M,$ preserving the metric tensor at the boundary.
We need to rule out examples like $[0,2\pi]\to S^1,$ where $[0,2\pi]$ get glued along $\{0\} \sim \{2\pi\}.$ I will argue that $f$ is a gluing map $M\to M/\pi\cong N$ where $\pi$ is a free diffeomorphic involution on a union of connected components of $\partial M.$ The map $\pi$ is defined by $\pi(x)=x'$ whenever $x,x'\in \partial M$ with $x\neq x'$ and $f(x)=f(x').$ We need to show this is well-defined and has a clopen domain.
If three distinct points $x_1,x_2,x_3$ were sent to the same point $y\in N,$ then $M^o\cap B(x_i,\epsilon)$ would be sent to, approximately, a half-sphere around $y.$ Three distinct such approximate half-spheres would have to intersect, a contradiction. So $\pi$ is well-defined.
Where two distinct points $x,x'$ are sent to the same point $y\in N,$ the point $y$ must be in the interior of $N.$ The neighborhoods of $x$ and $x'$ must end up boundary-to-boundary - the two half-spheres must completely fill the space near $y,$ because nothing else is going to. By the inverse function theorem the map $x\mapsto x'$ can locally be extended to a smooth map. And given a sequence of pairs of distinct $x_n,x'_n$ with $f(x_n)=f(x_n')$ and $x_n\to x,$ passing to a subsequence if necessary we get $x'_n\to x',$ and $x=x'$ is impossible because $f$ is locally injective (its derivative preserves the metric tensor, even at the boundary). So $\pi$ has a clopen domain.
This shows that $f$ is either injective hence an isometry, or it decreases the number of boundary components, which cannot happen if $M$ is diffeomorphic to $N.$
