Prove that this class of curves has constant speed and curvature

For the speed it is sufficient to remark that $$|\gamma'(s)|=\lim_{h\to0+}{|\gamma(s+h)-\gamma(s)|\over h}\ .$$ Here the RHS is independent of $s$, by assumption on $\gamma$.

For the curvature we have to extract the value of $\kappa(s)$ from distance measurements. We thereby may assume that $|\gamma'(s)|\equiv1$.

Consider the following model situation: $s\mapsto\gamma(s)=\bigl(x(s),y(s)\bigr)$ is a smooth arclength parametrized curve with $$x(0)=y(0)=0,\quad \dot x(0)=1,\quad \dot y(0)=0,\quad \theta(0)=0\ ,$$ where $\theta(s):={\rm arg}\bigl(\dot x(s),\dot y(s)\bigr)$ is the argument of the tangent direction. Then one has $$\eqalign{\dot x&=\cos\theta,\quad\ddot x=-\sin\theta\cdot\dot\theta,\quad x^{...}=-\cos\theta\cdot\dot\theta^2-\sin\theta\cdot\ddot\theta,\cr\dot y&=\sin\theta,\quad \ddot y=\cos\theta\cdot\dot\theta\cr}$$ identically in $s$, and therefore $$\ddot x(0)=0,\quad x^{...}(0)=-\kappa^2,\qquad \ddot y(0)=\kappa\ ,$$ where $\kappa:=\dot\theta(0)$ is the curvature of $\gamma$ at $(0,0)$. Taylor's theorem then gives $$x(s)=s-{\kappa^2\over6}s^3+O(s^4),\quad y(s)={\kappa\over2}s^2+O(s^3)\qquad(s\to0)\ .$$ It follows that $$|\gamma(s)|^2=\left(s^2-{\kappa^2\over3}s^4+O(s^5)\right)+\left({\kappa^2\over4}s^4+O(s^5)\right)=s^2-{\kappa^2\over12}s^4+O(s^5)\qquad(s\to0)\ ,$$ so that $$\lim_{s\to0}{|\gamma(s)|^2-s^2\over s^4}=-{\kappa^2\over12}\ .$$ Back to our given special curve $\gamma$ this means that $$-{\kappa^2(s)\over12}=\lim_{h\to0}{\bigl|\gamma(s+h)-\gamma(s)\bigr|^2-h^2\over h^4}\ .$$ Here the RHS is independent of $s$, hence $s\mapsto\kappa^2(s)$ is constant. By continuity this implies that $s\mapsto\kappa(s)$ is constant.