I'm studying harmonic analysis by myself. One of the online notes gives the following claim as a remark:

For any $N \in \mathbb{Z}^{+}$, let's use $S_{N}$ to denote the partial ($N$ terms) Fourier sum operator, i.e for any function $f$, the function $S_{N}f$ is given by: $$S_{N}f(x) = \sum_{n=-N}^{N}\hat{f}(x)e^{2\pi inx}$$ Then there exists some constant $C>0$, such that for any $f \in L^{1}([0,1])$ and any $\lambda > 0$, the following "analogous" weak-type (1,1) inequality holds: $$\sup_{N}\big|\big\{\theta \in [0,1] \ | \ |S_{N}f(\theta)| > \lambda\big\}\big| \leq \frac{C}{\lambda}\|f\|_{L^1}$$ where $|A|$ denotes the Lebesgue measure of a set $A$. Any ideas on how to prove this inequality? I have tried some approaches with some tricks used to prove the conventional Hardy-Littlewood weak-type maximal inequality, but it doesn't seem to work....Thanks in advance!

Moreover, the notes also says that this inequality is related to a weak form of the almost everywhere convergence of Fourier series on $L^1([0,1])$. More specifically speaking, we can use the weak-type inequality above to show that there exists a subsequence $\{N_{m}\}_{m=1}^{\infty}$ of $\mathbb{Z}^{+}$, such that $S_{N_{m}}f \rightarrow f$ pointwisely and almost everywhere as $m \rightarrow \infty$. Any idea on this?