I have to show that there is $f\in \mathcal C^0(S^1)$ s.t. $$\lim_{n\to \infty }\|S_nf-f\|_{L^\infty }\neq 0.$$ The proof goes as follow : we know that $\|D_n\|_{L^1}\geq c\log(n)$ where $D_n$ is the Dirichelet kernel. Therefore, $$\sup_{\|f\|_{L^\infty }\leq 1}|D_n*f(0)|=\|D_N\|_{L^1}\geq c\log(n).$$
I just don't understand why $$\sup_{\|f\|_{L^\infty }\leq 1}|D_n*f(0)|=\|D_N\|_{L^1}.$$
You've seen one inequality is obvious but we need the other direction. You can get this directly, constructing $f$ with $|f|\le1$ such that $\int fD_n$ is close to $||D_n||_1$. (Draw a picture of $D_n$ and imagine what $f$ should look like; you want $f=1$ on "most" of the set where $D_n>0$ and $f=-1$ on most of the set where $D_n<0$.)
Or you can get it from some fancy theorems. The Riesz Representation Theorem says that the norm of a complex measure $\mu$ is equal to the supremum of $|\int f\,d\mu|$ for $f$ continuous with $||f||_\infty\le1$, and something or other shows that the norm of $D_n$, regarded as a complex measure, is just $||D_n||_1$.