It is a well known result that for any compact, convex set $\Omega \subset \mathbb{R}^n$ and for any $\varepsilon > 0$ there are convex sets $\Omega_1 \subset \Omega \subset \Omega_2$ with C^2 boundaries $\partial \Omega_1$ and $\partial \Omega_2$ such that $d(\Omega, \partial \Omega_1) < \varepsilon$ and $d(\Omega, \partial \Omega_2) < \varepsilon$ where $d$ denotes the Hausdorff distance between sets (see for instance section 4.3 from Eggleston's Convexity).
It is also generally implied that convexity is a critical requirement and the above won't hold for non-convex domains. I'm trying to understand what fails in the non-convex case since (my) intuition seems to imply this should hold also for non-convex domains: a draw is far from a proof, but it looks like I could draw a regular (say $C^2$) set as close as desired to the non-convex set.
EDIT:
The set motivating me for asking this question is that made of a circle without a sector:
Let $\Omega = [-1,1]^2\setminus ((0,1]\times {0})$.
$\Omega$ is a closed square from whom we removed a line going to its center. You cannot construct a $C^2$ curve outside $\Omega$ going closer than $1$ to $0$ since the only way to achieve this is to use the line as you path. However, such curve cannot be differentiable everywhere because it must have a cusp.
Edit : please note that $\Omega$ is not compact in this example. So it does not fully answer the question. I am thinking about it.