Locally Lipschitz, boundary I have a question about domains of $\mathbb{R}^d$.
I am reading A First Course in Sobolev Spaces by Giovanni Leoni. 
This book introduces the following concept to describe the regularity of the boundary of a domain. 
(Definition 12.9 in Leoni's book) The boundary $\partial \Omega$ of an open set $\Omega \subset \mathbb{R}^N$ is $locally $ $Lipschitz$ if for each $x_0 \in \partial \Omega$ there exist a neighborhood $A$ of $x_0$, local coordinates $y=(y',y_N) \in \mathbb{R}^{N-1} \times \mathbb{R}$, with $y=0$ at $x=x_0$, a Lipschitz function $f:\mathbb{R}^{N-1} \to \mathbb{R}$, and $r>0$, such that
\begin{equation*}
\Omega \cap A=\{(y',y_N) \in \Omega \cap A: y' \in Q(0,r), y_N>f(y') \},
\end{equation*}
where $Q(0,r)$ is the open ball centered at the origin with radius $r>0$.
My question
In the above definition, $f$ and $r>0$ depend on $x_0$. Therefore, we should write $f=f_{x_0}$, $r=r_{x_0}$.
Can we show the following? 
Let $K \subset \partial \Omega$ be a compact subset. Then, $\inf_{x_0 \in K}r_{x_0}>0$ and there exists a positive number $M$ such that
\begin{equation*}
\sup_{x_0 \in K}\text{Lip}(f_{x_0})\le M,
\end{equation*}
where $\text{Lip}(f)$ is the Lipschitz constant of $f$.
I think this holds. But I couldn't prove... If you have a nice idea, please let me know.
Thanks in advance.
ADD
That is, I want to prove the following:
There exist positive numbers $r_K$ and $L_{K}$ such that
for each $x_0 \in K$ there exist a neighborhood $A$ of $x_0$, local coordinates $y=(y',y_N) \in \mathbb{R}^{N-1} \times \mathbb{R}$, with $y=0$ at $x=x_0$, a Lipschitz function $f:\mathbb{R}^{N-1} \to \mathbb{R}$ with $\text{Lip}(f) \le L_{K}$ and 
\begin{equation*}
\Omega \cap A=\{(y',y_N) \in \Omega \cap A: y' \in Q(0,r_K), y_N>f(y') \},
\end{equation*}
where $Q(0,r_K)$ is the open ball centered at the origin with radius $r_K>0$.
 A: For every point $x_0\in K$ you have an open neighborhood $A_{x_0}$ that contains $x_0$. Hence $K\subset\cup_{x_0\in K}A_{x_0}$ and so by compactness you can find $n_K$ points $x_1, \ldots, x_{n_K}\in K$ such that $K\subset\cup_{i=1}^{n_K} A_{x_i}$. So now take $L_K=\max_{i=1,\ldots,{n_K}} L_{x_i}$ and $r_K=\min_{i=1,\ldots,{n_K}}$.
If $x\in K$ then there exists $i\in\{1,\ldots,{n_K}\}$ such that $x\in A_{x_i}$.
EDIT Instead of working with cubes use balls (it does not make any difference). For each $x_0\in K$ find a ball $A_{x_0}=B(x_0,r_{x_0})$.  Hence $K\subset\cup_{x_0\in K}B(x_0,\frac{r_{x_0}}2)$ and so by compactness you can find $n_K$ points $x_1, \ldots, x_{n_K}\in K$ such that $K\subset\cup_{i=1}^{n_K} B(x_i,\frac{r_{x_i}}2)$. So now take $L_K=\max_{i=1,\ldots,{n_K}} L_{x_i}$ and $r_K=\min_{i=1,\ldots,{n_K}}\frac{r_{x_i}}2$.
If $x_0\in K$ then there exists $i\in\{1,\ldots,{n_K}\}$ such that $x_0\in B(x_i,\frac{r_{x_i}}2)$. Then $B(x_0,r_{K}) \subset B(x_i,r_{x_i})$. You can take as function $f$ the function $f_{x_i}$ but you will need to make a translation to change the origin into $x_0$.
You can use the same idea with cubes, it is just more cumbersome to write down.
