Point-wisely dense orthonormal basis

Let us denote $$T$$ by the unit circle. Let $$\{e_n\}$$ be an orthonormal basis for $$L^2(T)$$, with respect to Lebesgue measure.

We say $$\{e_n\}$$ is smooth if it satisfies the following property:

$$f(t){=}\lim_{N\to \infty}\sum_{-N}^{N}\langle f,e_n\rangle e_n(t)~~~~~(\forall f\in C(T))$$

Q. Does there exists any smooth orthonormal basis for $$L^2(T)$$?

Yes, you just need to find a Schauder basis of $$C(T)$$ that is an orthonormal basis of $$L^2(T)$$ at the same time. On the unit interval this can be done using the Franklin system, which has a periodic version, suitable for the unit circle. If you want the basis indexed by $$\mathbb{Z}$$ instead of $$\mathbb{N}$$, you can just reindex it. The coefficients of the decomposition in $$C(T)$$ are given by the inner products, since convergence in $$C(T)$$ implies convergence in $$L^2(T)$$, thus it has to be the usual orthonormal decomposition.