Does a map that preserves lengths of rectifiable curves preserve lengths of all curves? Let $X,Y$ be length spaces, and let $\, f:X \to Y$ be a map. 
Suppose that $f$ preserves the lengths of all rectifiable curves in $X$, i.e for every (continuous) path $\alpha$ such that $L_X(\alpha) < \infty$, $L_Y(f \circ \alpha)=L_X(\alpha)$.
Question: Is it true that $f$ preserves lengths of all paths, rectifiable or not?
i.e assume $L_X(\alpha) = \infty$. Is it true that $L_Y(f \circ \alpha)= \infty$?
In other words, does $f$ take non-rectifiable curves to non-rectifiable curves?
Does anything change if we assume $X,Y$ are Riemannian manifolds?

A naive approach would be as follows:
If $L_X(\alpha)=\infty$, then perhaps we can break $\alpha$ by restrictions $\alpha_n=\left.\alpha\right|_{I_n}$, such that $n \le L_X(\alpha_n) < \infty$, and so $L_Y(f \circ \alpha) \ge L_Y(f \circ \alpha_n)=L_X(\alpha_n) \ge n$. However, not every non-rectifiable curve can be broken in this way- think of the Koch's snowflake- every non-trivial restriction of it has an infinite length.
 A: On $M:=\mathbb{E}^3$, we have a metric
$$ g
=dx^2+dy^2+\omega^2,\ \omega=xdy-dz
$$ 
 Consider a projection $$\pi : (M,g)
 \rightarrow (\mathbb{E}^2,dx^2+dy^2),\ \pi (x,y,z)=(x,y)$$
Note that $$ d\pi \ V =(1,0),\ d\pi\ W=(0,1) $$
where $V=(1,0,0),\ W= (0,1,x)$ Since $V,\ W$ are orthonormal so
$d\pi$ is isometry on $(V,W)$
i) Define a new metric on $M$ : If $c(t)$ is a curve in $M$ from $p$
to $q$ s.t. $g(X,c'(t))=0$, where $\{ V,W,X\}$ is orthonormal frame
for $g$, then
$$d(p,q)={\rm inf}_c\ {\rm length}_g\ c$$
ii) $d(p,q) <\infty$ Here if the distance is obtained by $c$, then
${\rm length}_d\ c= d(p,q)={\rm length}\ \pi\circ c$
iii) $c(t)=(0,0,t)$ Note that $d(
c(t),c(t+\epsilon))=C\sqrt{\epsilon} $ for some $C$
So $c([0,1])$ has infinite length wrt $d$
[Updated]----
i) Define a new metric on $M$ :
If $c(t)$ is a piecewise smooth path in
$M$ from $p$ to $q$ s.t. $c'(t)\in (V,W)$ (We call it admissible
path) then
$$d(p,q)={\rm inf}_c\ {\rm length}_g\ c$$
Here $d(\cdot,\cdot)$ is Carnot-Caratheodory distance which gives to
$M$ structure of metric space Here $d$ is finite
ii) Every two pts can be connected by admissible
path (cf Chow-Rashevsky Connectivity Theorem)
iii) $G=\mathbb{E}^3$ acts on $M$ transitively and
isometrically :
$$
a=(x,y,z),\ a\cdot (x_1,y_1,z_1)=(x+x_1,y+y_1,z+z_1+xy_1) $$
Note that orthonormal vector fields $V,\ W$ are invariant under
$G$-action Hence action is isometric
iv) Ball-Box Theorem : $$P_r^c\subset B^d_r((0,0,0))
\subset P_r^C $$
where $B^d_r(p)$ is $r$-ball at $p$ wrt $d$ and $P_r^c$ is $$
[-cr,cr]\times [-cr,cr]\times [-cr^2,cr^2]\subset \mathbb{E}^3$$
By $iii)$ we have $x_k:=  (0,0,\frac{k}{n}),\
d(x_k,x_{k+1})=d(x_{k+1},x_{k+2})$ And Ball-Box theorem implies that
$$ d(x_0,x_1)=\frac{C_1}{\sqrt{n}}$$ for some $C_1$ Hence $$
\infty \leftarrow \sum_{k=1}^n d(x_k,x_{k+1}) < {\rm length}\ c$$
where $c : [0,1]\rightarrow M,\ c(t)=(0,0,t) $
Hence we have $$0={\rm length}\ \pi\circ c\neq {\rm length}\ c
=\infty$$
v) Note that by definition of $d$, smooth admissible path
is minimal geodesic in a small neighborhood By EXE 5.4.7, in fact
any two pt can be connected by smooth minimal geodesic
So since $d\pi$ is isometry $\pi$ preserves length of admissible
path 
vi) Assume that $c_n\rightarrow c$ in point wise, ${\rm
length}\ c<\infty$ and $c_n$ is admissible path
Note that $$ {\rm length}\ c=\lim\ {\rm length}\ c_n= \lim\ {\rm
length}\ \pi\circ c_n \geq {\rm length}\ \pi\circ c
$$ since $\pi\circ c_n\rightarrow\pi\circ c$ in point wise
If last inequality is strict assume that
$$c([a,b])\cap c_n([a,b]) =\{ c(a)=c_n(a), c(b)=c_n(b)\}$$
 and $c_n=\alpha_1\cup \alpha_2$ where $\alpha_1|[a,t],\ \alpha_2
|[t,b]$ are smooth admissible Here we change $\alpha_i$ with a
smooth minimal geodesic $\alpha_i'$ s.t. $$ \alpha_1(a)=
\alpha_1'(a),\ \alpha_2(b)=\alpha_2'(b),\
\alpha_1'(t)=\alpha_2'(t)=c(t) $$
From this infinitely many process we obtain equality
[Riemannian Case]
(1) If $c: [0,1]\rightarrow M$ is a curve then define a curve $c_n$
to be a broken geodesic joining $\{ c(\frac{k}{n} )\}_{k=0}^n $.
Hence $c$ is a pointwise limit of $c_n$.
(2) That is any curve $c$ is a limit of sequence of curves $c_n$ of
finite lengths. If $f$ is a weak path isometry, then $f$ is short.
Hence $f\circ c$ is a pointwise limit of $f\circ c_n$. Note that
$${\rm length}\ c=\lim\ {\rm length}\ c_n=\lim\ {\rm length}\
 f\circ c_n= {\rm length}\ f\circ c$$ 
A: I am filling some details in HK Lee's great answer.
We want to prove $\pi:(M,d) \to (\mathbb{E}^2,euc)$ preserves (finite) lentghs of paths. 
Take an arbitrary path $c$ in $M$, s.t $L_d(c) < \infty$. We claim that $L_e(\pi \circ c)=L_d(c)$. By the isomteric nature of $\pi$ it is enough to show that $c$ is consistent with the distribution $V,W$ (i.e $\dot c(t) \in \operatorname{span}(V(c(t)),W(c(t))  $ ) and that $L_d(c)=L_g(c)$.
Lemma 1:
$L_d(c) < \infty \Rightarrow \dot c(t) \in \operatorname{span}(V(c(t)),W(c(t))  $.
Assuming the lemma, we proceed as follows:
$L_g \le L_d$, since both $d,d_g$ are intrinsic and $d_g \le d$ (since in $d$ we allow less paths) so $L_g=L_{d_g} \le L_d$. 
The argument in the proof of proposition 2.3.12 in the book a course in metric geometry (Burago,Burago,Ivanov) shows that $L_{d_g} \ge L_d$.

Proof of lemma 1:
To complete...
