If $\Omega$ is a bounded domain in $\mathbf{R}^n$ with $C^2$ boundary, show that $\Omega$ satisfies "exterior sphere condition".
Exterior sphere condition means that for each $z\in\partial\Omega$, there is a ball $B_r(\xi)$ satisfying $\overline{B_r(\xi)}\cap\overline{\Omega}=\{z\}$.
Given a point on the boundary of $\Omega$, choose an orthogonal coordinate system $x_1,\ldots,x_n$ for $\mathbb{R}^n$ so that the point lies at $(0,0,\ldots,0)$ and the tangent hyperplane to $\partial\Omega$ at the point is the hyperplane $x_n = 0$.
In this case, by the Implicit Function Theorem, there is an $\epsilon$-neighborhood of the origin on which $\Omega$ is defined by $$ x_n \;<\; f(x_1,\ldots,x_{n-1}) $$ where $f\colon\mathbb{R}^{n-1}\to\mathbb{R}$ is a $C^2$ function and $f(0,\ldots,0)=0$. This function has a Hessian matrix $Hf$ at the origin, where $$ (Hf)_{ij} \;=\; \frac{\partial f}{\partial x_i\partial x_j}(0,\ldots,0). $$ This matrix is symmetric, and hence diagonalizable, with real eigenvalues $\lambda_1,\ldots,\lambda_n$. (These are the principal curvatures of $\partial\Omega$ at the origin.)
Fix a $C > \max\{\lambda_1,\ldots,\lambda_{n-1}\}$. Then there exists a $\delta>0$ so that $$ \|(x_1,\ldots,x_{n-1})\|<\delta \qquad\Rightarrow\qquad f(x_1,\ldots,x_{n-1}) \leq C(x_1^2 + \cdots + x_{n-1}^2) $$ Now, if $r > 0$ is any radius satisfying $1/r > 2C$, then the ball of radius $r$ centered at $(0,\ldots,0,r)$ lies entirely in the region $x_n \geq C(x_1^2+\cdots + x_{n-1}^2)$, and intersects the boundary $x_n=C(x_1^2+\cdots +x_{n-1}^2)$ only at the origin. It follows that if $0 < r < \min\{\delta,1/(2C)\}$, then this ball will intersect $\overline{\Omega}$ only at the origin.
