Let $\Omega$ be a domain in $\mathbb{R}^n$. Then I can find a bounded subdomain $\Omega_0$ in $\Omega$ with $r=\mathrm{dist}(\Omega_0,\partial\Omega)>0$. Here domain means an open and connected subset of $\mathbb{R}^n$.
I want to find a bounded Lipschitz domain $\Omega_1$ in $\mathbb{R}^n$ which contains $\Omega_0$ and contained in $\Omega$.
I have a rough idea about the proof, but I'm not satisfying this strategy because I cannot make a rigorous proof based on this idea.
Cover $\Omega_0 \cup \partial \Omega_0$ by a family of open cubes whose diameter is less than $r$. Then by compactness, there exists a finite family $\mathcal{F}$ of cubes which covers $\Omega_0 \cup \partial \Omega_0$. Enlarge each $Q$ in $\mathcal{F}$ so that for any $Q\in \mathcal{F}$, there exists $Q'\in\mathcal{F}$ with $Q\neq Q^\prime$ such that $Q\cap Q' \neq \varnothing$.
Define $\Omega_1$ be the union of the above cubes. Then $\Omega_1$ is clearly open and connected. So it is a domain in $\mathbb{R}^n$.
Is union of two Lipschitz domains (with nonemtpy intersection) is Lipschitz? If so, then since cubes are Lipschitz domain and $\Omega_1$ is a finite union of Lipschitz domains, so I can prove the claim.
It seems obvious in 2D by picture but I cannot make any progress. One of my trial was changing parameter when two domain meets.
I can find such a domain with $C^1$-boundary using Sard's lemma. So the existence is clear. But I want to find different method based on elementary arguments.
Thanks for help.
It is not true. Take two squares in $\mathbb{R^2}$ that touch only at one corner, say $Q_1=(0,1)\times (0,1)$ and $Q_2=(1,2)\times (1,2)$. Then each cube is Lipschitz but the union is not, since at the point $(1,1)$ you cannot write the boundary as the graph of a function.