Is it true that $\mathbb{R}^n_+=\mathbb{R}^{n-1}\times (0,\infty)$ is trivially $(\varepsilon, \delta)$-locally uniform, for $\varepsilon=1$ and any $\delta$ and the arc chosen as the segment of a straight line connecting $x$ and $y$?
The definition:
A domain $\Omega \subseteq \mathbb{R}^n$ is $(\varepsilon, \delta)$-locally uniform for $\varepsilon, \delta>0$ if any $x,y \in \Omega$ with $\| x-y \| \leq \delta$ can be joined by a rectifiable arc $\gamma \subset \Omega$ satisfying for any point $z \in \gamma$
$l(\gamma) \leq \varepsilon \| x-y \|$, and
$\min\{l(\gamma'), l(\gamma'')\} \leq \mathrm{dist}(z, \partial \Omega)$,
where $l(\gamma)$ is the Euclidean length of the arc $\gamma$, and $\gamma'$ and $\gamma''$ are the two components of $\gamma\setminus \{z\}$. $\partial \Omega$ is the boundary of $\Omega$.
I am only concerned about defining $\partial \Omega$ for an unbounded space.
P.S. if this holds, $\mathbb{R}^n_+$ is uniform as well (drop the "locally"), ie $\delta=\infty$; the same should hold for any convex set.