I found the following (obviously mistaken) proof that any continuous function $f\in C[\mathbb{T}]$ for which $|{f-\mathcal{S}_{N}f}|_{\infty}\to0$ must be smooth.
Can you please help me find the flaw?
The 'proof':
Suppose that $|{f-\mathcal{S}_{N}f}|_{\infty}\to0$, i.e., the Laurent series $\sum_{n\in\mathbb{Z}}\hat{f}(n)z^{n}$ converges on $\partial\mathbb{D}$. In particular: $f(t)=\sum_{n\in\mathbb{Z}}\hat{f}(n)\left(e^{2\pi it}\right)^{n}$ on $\mathbb{T}$. As a Laurent series must either converge locally uniformly on an open set or diverge everywhere, there is an open neighborhood $\partial\mathbb{D}\subseteq U$ on which the series $\sum_{n\in\mathbb{Z}}\hat{f}(n)z^{n}$ converges locally uniformly, say to a function $F:U\to\mathbb{C}$.
Now take a point $w\in U\backslash\left\{ 0\right\}$ . As $\sum_{n\in\mathbb{Z}}\hat{f}(n)z^{n}$ converges locally uniformly on U, there is $0<\epsilon<\left|w\right|$ for which the series converges uniformly on $B_{\epsilon}(w)$. Let $\gamma$ be any closed piecewise $C^{1}$ loop in $B_{\epsilon}(w)$. As $\sum_{n=-N}^{N}\hat{f}(n)z^{n}$ is analytic on $B_{\epsilon}(w)$ (because it's only pole, $0$, is outside of $B_{\epsilon}(w)$), By Cauchy's Theorem $\int_{\gamma}\sum_{n=-N}^{N}\hat{f}(n)z^{n}\text{ d}z=0$ for every $N\in\mathbb{N}$. As the convergance on $B_{\epsilon}(w)\subseteq U$ is uniform, by the theorem about swapping a limit with an integral we get:$$\int_{\gamma}F(z)\text{ d}z=\int_{\gamma}\sum_{n\in\mathbb{Z}}\hat{f}(n)z^{n}\text{ d}z=\int_{\gamma}\lim_{N\to\infty}\sum_{n=-N}^{N}\hat{f}(n)z^{n}(z)\text{ d}z=$$ $$=\lim_{N\to\infty}\int_{\gamma}\sum_{n=-N}^{N}\hat{f}(n)z^{n}(z)\text{ d}z=\lim_{N\to\infty}0=0$$The choice of $\gamma$ was arbitrary, hence by Morera's theorem $F$ is analytic on $B_{\epsilon}(w)$, and in particular on $w$. The choice $w$ was arbitrary as well, hence $F$ is analytic on $U\backslash\left\{ 0\right\}$ which is an open neighborhood of $\partial\mathbb{D}$. As mentioned before, for every $t\in\mathbb{T}$ we have: $$f(t)=\mathcal{S} f(t)=\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi int}=F\left(e^{2\pi it}\right)$$ Thus $f$ is smooth.