Let $U$ be an open set in $\mathbb{R}^n$, let $E$ be the set of norms on $\mathbb{R}^n$, and let $N: U\rightarrow E$ be a map such that $(x,v) \mapsto N(x)(v)$ is continuous.
We define the length of a piecewise-$C^1$ curve $\gamma : [0,1] \rightarrow U$ to be $L(\gamma) := \int^1_0 N(\gamma(t)) (\gamma'(t)) \mathbf{d}t$.
We then define the distance $d(x,y)$ between two points $x$ and $y$ to be the infimum of the lengths of curves joining $x$ and $y$. Obviously, piecewise-$C^1$ curves are $d$-rectifiable. However, I have struggle trying to prove that the length in the sense of rectifiable curves is equal to the length defined above.
Indeed, for all subdivision $0 = t_0 < t_1 < \cdots < t_n < 1$, $\sum^{n-1}_{i=0} d(\gamma(t_{i+1}), \gamma(t_i)) \leq L(\gamma)$, because for each $i$, $d(\gamma(t_{i+1}),\gamma(t_i)) \leq L(\gamma_{|_{[t_i, t_{i+1}]}})$ and this proves rectifiability.
I managed to show that $\int^1_0 N(\gamma(t)) (\gamma'(t)) \mathbf{d}t$ and $\sum^{n-1}_{i=0} \int^{t_{i+1}}_{t_i} N(\gamma(t_i))(\gamma'(t)) \mathbf{d}t$ are close, if the subdivision is very thin; then $\sum^{n-1}_{i=0} \int^{t_{i+1}}_{t_i} N(\gamma(t_i))(\gamma'(t)) \mathbf{d}t$ is close to $\sum^{n-1}_{i=0} \int^{t_{i+1}}_{t_i} \frac{1}{t_{i+1}-t_i}N(\gamma(t_i))(\gamma(t_{i+1}) - \gamma(t_i)) \mathbf{d}t$ but in general, segments are not geodesics for $d$, I get stuck.