It depends on what is meant by "polynomial".
If only $\sum c_n z^n$, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of $J$.
Although that condition is trivially satisfied if $J$ has empty interior, that doesn't mean that for such $J$ every continuous function is the uniform limit of polynomials. For example the unit circle has empty interior, but a sequence of polynomials converging uniformly on the unit circle converges uniformly on the closed unit disk by the maximum principle, and thus if $f$ is a uniform limit of polynomials on the unit circle, then there is a holomorphic function $h$ on the unit disk that extends continuously to the unit circle, with boundary values $f$. In particular, we have
$$\int_{\lvert z\rvert = 1} f(z)\cdot z^n \,dz = 0\tag{1}$$
for all $n \geqslant 0$. (And, in this case, that condition is sufficient.)
That phenomenon generalises, if $J$ disconnects the plane, that is, if $\mathbb{C}\setminus J$ has at least two connected components, then the bounded components of the complement of $J$ impose restrictive conditions on the continuous functions that are uniform limits of polynomials similar to $(1)$.
Mergelyan's theorem asserts the converse, if $J$ is a compact subset of $\mathbb{C}$ with empty interior such that $\mathbb{C}\setminus J$ is connected, then every continuous function on $J$ can be uniformly approximated by polynomials (in $z$ only).
If "polynomial" means polynomial in $z$ and $\overline{z}$, or equivalently polynomial in $\operatorname{Re} z$ and $\operatorname{Im} z$, then the Weierstraß approximation theorem holds for all compact $J$.
The hypotheses of Weierstrass satisfy the hypotheses of Mergelyan:
$K=[a,b] \subseteq \mathbb C$ is a compact set with connected complement.
Since $K$ has empty interior, "holomorphic in the interior of $K$" is true.
The conclusions of Mergelyan imply the conclusions of Weierstrass:
"approximated uniformly" is the same as "$| f(x)− p(x)| < \epsilon$ for all $x \in K$".
Best Answer
No, because the set of polynomials in z is not self conjugate. If you have a series of polynomials in z that converges in the supremum norm on D, the limit function needs to be holomorphic again, showing that an arbitrary continuous function cannot be approximated by such polynomials: All continuous functions would need to be holomorphic.
In order to apply the Stone-Weierstrass theorem, you'd need to consider polynomials in z and $\bar z$. Let $h(z)$ be a continuous function. Then we can write $$ h(z) = f(z) + i g(z) $$ with real valued functions f and g. These can be approximated as real valued functions with polynomials $p_f (x, y)$ and $p_g(x, y)$ in $x, y$ by the real version of the Stone-Weierstrass theorem. Any polynomial in $x, y$ can be transformed into a polynomial in the variables z and $\bar z$, so that we can make $$ \| h(z) - (p_f(z, \bar z) + i p_g(z, \bar z)) \|_{\sup} $$ arbitrarily small.
(This is the proof of the complex version using the real version of the Stone-Weierstrass theorem given by Lang applied to this concrete situation.)
Since the Stone-Weierstrass theorem is essentially about continuous functions and not about holomorphic ones, I doubt that there are any neat tricks from complex calculus that could make the general proof given by Lang easier, shorter, more elegant in this particular case.
Note: We are referring to the book