I think the definition of length metric works without assuming curves are continuous.
Let $X$ be a subset of $\mathbb{R}^n$ and $x, y\in X$.
Def 1. For a function $f:[0, 1]\rightarrow X$, $L(f)$ is defined by the sup of the length of polygonal lines associated with a partition of $[0, 1]$. We call $f$ is rectifiable if $L(f)<\infty$.
Assumption: For all $x, y\in X$, there is a rectifiable function $f:[0, 1]\rightarrow X$ such that $f(0)=x, f(1)=y$.
Def 2. For $x, y\in X$, the distance $d(x, y)$ is defined to be the inf of $L(f)$ for rectifieable functions $f:[0, 1]\rightarrow X$ such that $f(0)=x, f(1)=y$.
I think this $d$ satisfies the axiom of metric without assuming continuity of $f$.
Is it right?
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Prop $d(x, y)=0\Leftrightarrow x=y$
Proof. (1) $\Leftarrow$ is obvious. (2) If $d(x, y)=0$, for all $\varepsilon>0$, there is a function $f:[0, 1]\rightarrow X$ such that $f(0)=x, f(1)=y, L(f)<\varepsilon$. There is a partition $t_1=0\leq t_1\leq\dots\leq t_n=1$ such that the length $L$ of the polygonal line connecting $f(t_1), \dots, f(t_n)$ is less than $\varepsilon$. So the length of the segment connecting $f(0)$ and $f(1)$ is less than $\varepsilon$. Thus $x=y$.
Yes, you can drop the assumption about $f$ being continuous and you will still get a metric. However this won't give you anything interesting. That's because in such scenario, the newely defined metric is just the Euclidean one.
First of all given $f:[0,1]\to X$ (continuous or not), with $f(0)=x$ and $f(1)=y$ we have that $L(f)\geq \lVert x-y\rVert$. That's because of the nature of the Euclidean space: the shortest polygonal chain (or any path) between any two points is just a straight line. Now all you need to realize is that the $L(f)=\lVert x-y\rVert$ equality is always doable if $f$ is not assumed to be continuous. By $f([0,1))=x$ and $f(1)=y$. And therefore $d(x,y)=\lVert x-y\rVert$ has to hold.