Suppose $X$ is a metric space with distance function $d\colon X\times X\to \mathbb{R}$. It is stated here that if a continuous curve $\alpha\colon [0,1]\to X$ has finite length, then it admits a Lipschitz parametrization, i.e., there exists a homeomorphism $g\colon [0,1]\to [0,1]$ such that $\alpha \circ g$ is a Lipschitz map. I'm interested in finding a proof for this result.
So far, I've been able to prove a weaker version of this statement: Let $L=\mathcal{L}_{0}^{1}(\alpha)$ be the total length of $\alpha$. The function $g\colon [0,1]\to [0,L]$ given by $g(s)=\mathcal{L}_{0}^{s}(\alpha)$ is a surjective, nondecreasing continuous function. If we assume that $\alpha$ is nonconstant in every subinterval $[a,b]\subseteq [0,1]$, then $g$ is a homeomorphism, and $\alpha \circ g^{-1}\colon [0,L]\to X$ is a unit speed parametrization of $\alpha$, hence Lipschitz. If we don't assume this, at least it's possible to see that $\alpha$ passes continuously to a $1$-Lipschitz function $\beta\colon [0,L]\to X$ such that $\alpha = \beta \circ g$, which is not quite the same as what's stated in the result, since $g$ is not a homeomorphism.
Can anyone give me a reference or an outline of the proof? Thanks in advance!
Let $\alpha$ be your rectifiable curve of length $L$, connecting point $p$ to a point $q$. Given $s\in [0, L]$, there is a unique point $x\in \alpha$ such that the 1-dimensional measure of $\alpha$ between $p$ and $x$ equals $s$. The map $f: s\mapsto x$ is called the arc-length parameterization of $\alpha$. The map $f$ is a homeomorphism and is 1-Lipschitz.