Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis of $L^2(S^1)$ with Haar measure Prove $\{z^n|n \in \mathbb{Z}\}$ is an orthonormal basis.

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*We show that this is an orthonormal set.

All we need to show is that $\int_{S^1} z^n d\mu=0$ since $\mu$ is a Haar measure $\int_{S^1} z^n d\mu=\int_{S^1} (az)^n d\mu=a^n\int_{S^1} z^n d\mu$, but if $a$ is not a root of unity, this implies the integral is $0$. Thus the set is orthonormal.


*We show the  the span is dense in $C(S^1)$ which in turn is dense in $L^2(S^1)$ which proves it is an orthonormal basis.

The Span is clearly an algebra so all we have to show is that it separates points. But that is fairly obvious. $x\not = y$ then take the identity map.
Is this correct?

You missed another condition. You need the fact that the span contains the complex conjugate of each of its elements. For this note that the conjugate of $z^{n}$ is $\frac 1 {z^{n}}=z^{-n}$ when $|z|=1$.
