A domain $\Omega \subset \mathbb{R}^n$ is called an $(\varepsilon, \delta)$-domain (or locally uniform domain) if there exists an $\varepsilon > 0$ such that for all $x, y \in \Omega$ with $\lvert{x - y}\rvert < \delta$ there is a rectifiable curve $\gamma : [0, l(\gamma)] \to \Omega$ joining $x$ to $y$ and satisfying \begin{align} l(\gamma) &\leq \frac{1}{\varepsilon} \lvert{x - y}\rvert,\text{ and} \label{eq:1}\tag{1}\\ \operatorname{dist}(z, \partial{\Omega}) &\geq \frac{\varepsilon \lvert{x - z}\rvert \lvert{y - z}\rvert}{\lvert{x - y}\rvert} \quad \forall z \in \gamma,\label{eq:2}\tag{2} \end{align} where $l(\gamma)$ denotes the arclength of $\gamma$.
These domains where introduced by Jones where he states (in the above notation)
Condition \eqref{eq:1} says that $\Omega$ is locally connected in some quantitative sense. Condition \eqref{eq:2} says there is a "tube" $T, \gamma \subset T \subset \Omega$; the width of $T$ at a point $z$ is on the order of $\min (\lvert{x-z}\rvert, \lvert{y-z})$. It is clear that every Lipschitz domain is an $(\varepsilon, \delta) $-domain for some values of $\varepsilon,\delta >0$.
Why is every Lipschitz domain an $(\varepsilon, \delta)$-domain?
What I've done:
I think this can only work on bounded domains as the strip $0 < x_2 < 1$ in $\mathbb{R}^2$ is Lipschitz but not uniform. Therefore, I'll assume $\Omega$ to be bounded. It is enough to prove the conditions for points near the boundary as otherwise a straight line suffices. Lipschitz domains satisfy the cone condition and this might help to find the curve. Now I am not sure whether I should explicitly construct the curve with line segments which seems quite tedious or if there is another way to state its existence.
Not a complete proof, but maybe a first step: I would use the definition of Lipschitz boundary, i.e. that for any $p\in\partial\Omega$ exists a radius $\delta>0$ and a bilipschitz function (i.e. bijective Lipschitz function, such that the inverse is Lipschitz as well) $\psi: B_\delta(p)\rightarrow B_1(0)$ such that $$\psi|_{\partial \Omega\cap B_\delta(p)} :\partial \Omega\cap B_\delta(p)\rightarrow B_1(0)\cap (\mathbb{R}^{n-1}\times\{0\}) $$ is a bilipschitz map. Furthermore $$\psi(\Omega\cap B_\delta(p)) = B_1(0)\cap \{(x_1,\ldots, x_n)|\ x_n>0\}.$$ Then for $x,y\in \Omega\cap B_\delta(p)$ we can connect $\psi(x)$ with $\psi(y)$ with a straight line $L:[0,1]\rightarrow B_1(0)$. Then the curve $$\gamma (t) := \psi^{-1}(L (t))$$ is rectifiable and after reparametrisation probably satisfies your requirements. But this is something you should check for yourself.
One other point: For this to work, $\overline{\Omega}$ probably has to be compact.