I am trying to prove what seems to be a standard result on the convergence of Fourier series,namely that if $ f:[-\pi,\pi] \longrightarrow \mathbb{C}$ is a continuous function then $$S_n(f)=\sum_{-n}^{n}\widehat{f}(k)e^{iks}$$ converges almost everywhere to $f$.
Here is my attempt: First we have to justify that $S_n(f)$ converges to something meaningfull . Since $ |\widehat{f}(k)|=|\frac{1}{2\pi}\int f(t)e^{-ikt}dt|\leq \frac{1}{2\pi}\int |f(t)|e^{-ikt}dt \leq \Vert f\Vert_{\infty}C\frac{1}{2\pi k}$ for some constant $C$,it is clear that the coefficients are square summable and we can use the weierstarss criterion. The value of the limit should coincide with $f$ because Fejer's theorem holds $\sigma_{n}(f)\longrightarrow f$ for continous functions and $ S_{n}(f) $ cannot tend somewhere else.
But why almost everywhere convergence?? What am i missing?
Your bound on $\widehat f(k)$ is incorrect. The correct one is $$|\widehat f(k)|\le\frac{1}{2\,\pi}\int_{-\pi}^\pi|f(t)|\,dt\le\|f\|_\infty.$$
Since $f$ is continuous, $f\in L^2$. Carleson's theorem gives the result.