Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?

Assuming "polynomial" means "polynomial in $z$", i.e. of the form $\sum_{k=0}^n a_k z^k$, then no, they are not dense, and the function $\bar{z}$ is not in their uniform closure.

The high-level reason is that polynomials are holomorphic on the unit disk, holomorphic functions are closed under uniform convergence, and $\bar{z}$ is not holomorphic.

For a direct proof, consider the linear functional $I(f) = \int_{S^1} f(z)\,dz = \int_0^{2\pi} f(e^{i\theta}) i e^{i\theta}\,d\theta$. This is clearly continuous with respect to the uniform norm since $$|I(f)| \le \int_0^{2\pi} |f(e^{i\theta})| |ie^{i\theta}|\,d\theta \le 2 \pi \|f\|_{\infty}.$$

But for any polynomial $f$, you can check that $I(f)=0$. Hence by continuity, $I(f)=0$ for any $f$ in the closure of the polynomials. On the other hand, for $f(z)=\bar{z}$, $I(f) = 2\pi i$, so $f(z)=\bar{z}$ is not in the closure.

(This proof is easy to discover when you know the Cauchy integral theorem, which says that any holomorphic function has $\int_{S^1} f(z)\,dz = 0$, or indeed the same replacing $S^1$ with any nice closed curve.)

Ok (almost) every periodic function $g$ on $[0,1]$ can be written in the form $$ g(x) = \sum_{n=-\infty}^\infty c_n e^{2\pi i n x} $$

You can switch from function $f : S^1 \subset \mathbb{C} \rightarrow \mathbb{C}$ to the periodic function on $[0,1]$ via simple transformation: $e^{2\pi i \theta}$. $$ g(\theta) = f(e^{2\pi i \theta}) $$ This transformation can be seen in some sense as bijection between periodic functions and functions defined on $S^1$. (You would need to exactly specify those function spaces to be true.)

But let's have a look what you get when you transform polynomial. $$ f(z) = \sum_{n=0}^N a_n z^n $$ $$ g(\theta) = f(e^{2\pi i \theta}) = \sum_{n=0}^N a_n e^{2\pi i n \theta} $$ Uupss you are missing those negative terms $e^{-2 \pi i n \theta}$. So you can't get every periodic function on $[0,1]$.

From above you can see that "polynomials" of form $\sum_{n=-N}^n a_n z^n$ are dense in $C(S^1,\mathbb{C})$ and they are holomorphic in $\mathbb{C} \setminus \{0\}$, which is interesting because $\overline z$ is not holomorphic and $\overline z$ was exactly missing in previous. This seams to me as very special to the $S^1$ because $\frac1{z} = \frac{\overline z}{z\overline z} = \overline z $ only on the $S^1$.

Could anyone expand on the thing that harmonic polynomials form basis of $L^2(S^1)$ and real part of holomorphic function is harmonic I guess it is connected to the above but right now I do not have the power to think any more.