Suppose $E(closed)\subset\{z:|z|=1\}$ and let $f(z)$ be a continuous function on the set $E$. I want to show that $f(z)$ can be approximated by polynomials on $E$.
I am not exactly sure how to solve this problem, but I have a very small idea. I know that Runge's theorem cannot apply because I am not told that the function is analytic so I would have to apply the (Stone) Weierstrass theorem.
Let $f(z) = \bar {z}.$ Claim: $\sup_{|z|=1}|f(z)-p(z)| \ge 1$ for all holomorphic polynomials $p.$ Proof:
$$\int_0^{2\pi} |f(e^{it}) - p(e^{it})|^2 dt= \int_0^{2\pi} |e^{-it} - p(e^{it})|^2 dt=\int_0^{2\pi} |1 - e^{it}p(e^{it})|^2 dt \ge 2\pi.$$
by simple orthogonality. But the latter integral is $\le 2\pi (\sup_{|z|=1}|f(z)-p(z)|)^2,$ and that proves the claim.