$u\in C^1(\overline{A}),0<u(x)\ \in A,u(x)=0,\partial u/\partial\nu(x)<0\ on\ \partial A$ imply $\exists e>0$ so $e\mbox{dist}(x,\partial A)\leq u(x)$

$A$ is a bounded domain in $\mathbb{R}^N$ with $\partial A$ of class $C^2$

$u:\overline{A}\to\mathbb{R}$ - we write $0\leq u$ if $0\leq u(x)$ for a.e. $x\in A$ and $0<u$ if $0\leq u$ and $0<u(x)$ in a subset of $A$ having positive measure.

Show that $u\in C^1(\overline{A})$, $u(x)>0$ for every $x\in A$, $u(x)=0$ on $\partial A$ and $\frac{\partial u}{\partial \nu}(x)<0$ ($x\in \partial A$ and $\nu=\nu(x)$ denotes the unit outer normal to $A$ at $x$), IMPLY there is some $\varepsilon>0$ such that $\varepsilon\mbox{dist}(x,\partial A)\leq u(x)$ for all $x\in\overline{A}$

Suppose by contradiction that $\forall \varepsilon>0,\ \exists x_\varepsilon\in \overline{A}$ s.t. $\varepsilon\mbox{dist}(x_\varepsilon,\partial A)> u(x_\varepsilon)$. Clearly, $x_\varepsilon \in A$. Let $\varepsilon=1/n$ and on has $\frac{1}{n}\mbox{dist}(x_n,\partial A)> u(x_n)$. If we suppose now that $x_n\to x_0\in\overline{A}$, then $0\geq u(x_0)$. If $x_0\in A$ then we have a contradiction with $u(x_0)>0$. So, $x_0\in\partial A$, $u(x_0)=0$ and $\frac{\partial u}{\partial \nu}(x_0)<0$.

*** Now, how I can obtain a contradiction?