$\textbf{Definition}$: A domain $\Omega\subset\mathbb{R}^n$ is said to be of class $C^{0,1}$ if for every $x\in\partial\Omega$ there exist $r>0$ and Lipschitz continuous function $\gamma:\mathbb{R}^{n-1}\rightarrow\mathbb{R}$ such that (may change the index of axis if necessary)$$\Omega\cap B_r(x_0)=\{x\in B_r(x_0):x_n>\gamma(x_1,...,x_{n-1})\}$$
$\textbf{Definition}$: A domain $\Omega\subset\mathbb{R}^n$ is said to satisfy the "segment condition" if for every $x\in\partial\Omega$ there exist an open ball $U_x$ containing $x$ and a vector $y_x\neq 0$ such that for each $z\in\overline{\Omega}\cap U_x$ we have $$\{z+ty_x:t\in (0,1)\}\subset\Omega$$
$\textbf{Goal}$: Prove that a $C^{0,1}$ domain satisfy the segment condition.
$\textbf{My attempt}$: For $\Omega\in C^{0,1}$ choose $U_x=B_{r/2}(x)$ and $y_x=\frac{1}{3}e_n$ where $e_n$ is the n-th unit vector. From the continuity of $\gamma$ it is clear that for each $z\in\overline{\Omega}\cap U_x$ we have $z_n\geq \gamma(z_1,...,z_{n-1})$ and therefore $\{z+ty_x:t\in (0,1)\}\subset\Omega\cap B_r(x_0)\subset\Omega$.
$\textbf{Question}$: Lipschitz condition is superfluous in my proof to the statement. Is Lipschitz condition necessary for a boundary to satisfy segment condition?
Any help would be appreciated.
A Lipschitz condition is not necessary for the segment condition to hold; for instance take $\Omega=\{ (a,b)\in \mathbb{R}^2: \sqrt{|a|}<b\}$, then it clearly satisfies the segment condition with $y_x=(0,1)$ for any $x=(a,b)$. If instead of segments you have cones in a fixed direction and aperture, then this new condition is equivalent to Lipschitz (it's called the interior cone condition).
There is some subtlety here though: Your proof at first glance may suggest that any graph (or rather the domain above said graph, i.e. $\Omega=\{ (x,t)\in \mathbb{R}^n: f(x)<t\}$) will satisfy this segment condition. However, consider for instance $f(x)= \sin(1/x)$ if $x\neq 0$ and $f(0)=0$. Then $(0,-1)\in \bar{\Omega}$, but any segment starting from this point will intersect the complement of $\Omega$. If your function $f$ is continuous though, then your proof goes through, owing to the fact that $\partial \Omega$ is exactly the graph of $f$ in this case.