Let $A$ be the set of all unit-length arcs in $\mathbb{R}^2$. Let $C_0, C_1 \in A$. My goal is to find a continuous function $f: [0,1] \rightarrow A$ such that $f(0)=C_0$ and $f(1)=C_1$, but I do not have a metric defined on $A$, nor do I have a precise way of identifying a particular arc in $A$. Maybe a good way around these barriers is to identify each arc with a parametric equation for that arc.
Assume we have parametric equations for $C_0$ and $C_1$, i.e. we know continuous functions $c_0: [0,1] \rightarrow \mathbb{R}^2$ and $c_1: [0,1] \rightarrow \mathbb{R}^2$ whose images are $C_0$ and $C_1$, respectively. It is sufficient to find a continuous function $g$ whose domain is $[0,1]$, whose codomain is the set of parametric equations for elements of $A$, and where $g(0) = c_0$ and $g(1) = c_1$. Now we may use what I believe to be a natural metric on these parametric equations: \begin{align*} d(c,c^*)= \sup_{t\in [0,1]}d'(c(t),c^*(t)) \end{align*} where $d'$ is the Euclidean metric for $\mathbb{R}^2$.
My first idea was to simply choose $\,g(s)=(1-s)c_0 +s\,c_1$, a continuous function satisfying $g(0) = c_0$ and $g(1) = c_1$, but it doesn't quite work because there is no guarantee that all parametric equations in the range of this function actually graph an arc of unit length. Essentially I need a function which preserves the arclength of our arcs as it deforms them, and I'm a bit stuck on how to approach this.
I apologize if I have over-complicated this problem or misused any language or anything; I have never had a course in topology and this is basically a question I came up with for myself.
Using your notation let $L(s)=\operatorname{Length}(g(s))$. Then $L(0)=L(1)=1$. Does $$ f(s)=\frac1{L(s)}\,g(s) $$ satisfy your requirements?
The only problem with the above approach would be if $L(s)=0$ for some $s\in(0,1)$. However this can't happen only if $c_1$ and $c_2$ are translates one of the other and have opposite orientation. For sim`plicity assume that $C_i$ is $C^1$. In that case $$ \operatorname{Length}(c_i)=\int_0^1\|c_i'(t)\|\,dt=1,\quad i=1,2, $$ and $$ L(s)=\int_0^1\|(1-s)c_1'(t)+s\,c_2'(t)\|\,dt. $$ If $L(s)=0$ then $$ (1-s)c_1'(t)+s\,c_2'(t)=0\implies c'_2(t)=-\frac{1-s}{s}\,c'_1(t)\quad\forall t\in[0,1]. $$ Taking norms and integrating in $[0,1]$ gives $$ \frac{1-s}{s}=1\implies s=\frac12, $$ from where $c'_2()t)=-c´(1(t)$ follows.