Let $\Omega$ be an open and bounded subset of $\mathbb{R}^2$ and let $C^k(\Omega)$, $1\leq k<\infty$, be the space of functions $f$ with continuous derivatives of order $\leq k$ in $\Omega$, endowed with the usual topology of semi-norms
$$|f|_{K}=\sup_{|p|\leq k}~\sup_{x\in K}|(\partial/\partial x)^p f(x)|,$$
where $K$ runs over the compact subsets of $\Omega$. Then, the polynomials are dense in $C^k(\Omega)$, see e.g. Treves, Topological vector spaces, distributions and kernels.

Let us now consider the subset $C^k(\overline\Omega)\subset C^k(\Omega)$ of functions whose derivatives extend continuously up to the boundary of $\Omega$, endowed with the topology of the norm
$$\|f\|_{\overline\Omega}=\sup_{|p|\leq k}~\sup_{x\in\overline\Omega}|(\partial/\partial x)^p f(x)|.$$
Is it true that the polynomials are dense in $C^k(\overline\Omega)$ ?

By Whitney extension theorem, if $\Omega$ has some smoothness, like quasi-convexity, functions of $C^k(\overline\Omega)$ can be extended, which implies density of polynomials, so my question is about general $\Omega$.

I would appreciate any reference, counter-example,...

I am also interested in the case of $C^\infty(\overline\Omega)$ defined as the intersection of all $C^k(\overline\Omega)$.

To answer the comment by ACL, here is an exemple of a domain $\Omega\subset\mathbb{R}^{2}$ and a function $f\in C^{1}(\overline\Omega)$ which cannot be extended to any neighborhood of $\overline\Omega$ :

Define the domain $\Omega$ to be the square centered at the origin, of length 2, from which a spine delimited by $y=\pm e^{-1/x}$, $x\geq 0$, has been removed. The function $f$ is the zero function except on the first quadrant where $f(x,y)=x^{2}$. Then $f$ is $C^{1}$, and $f$ and its derivatives of order 1 admit continuous limits on the boundary of $\Omega$, hence, by definition, $f\in C^{1}(\overline\Omega)$. If $f$ could be continued to a $C^{1}$ function on a neighborhood of $\overline\Omega$, it would be Lipschitz at 0, which is not the case since there is no constant $C$ such that $x^{2}\leq 2Ce^{-1/x}$ when $x\to 0$.