$S_N(f):=\sum\limits_{|n|\le N}\hat f_n\cdot e^{i2\pi nx}$ where $\hat f_n=\int_0^1f(x)e^{-i2\pi nx} dx$ and $f$ a $1$-periodic $C^1(\mathbb R,\mathbb C)$ function, then $S_N(f)$ converges pointwise to $f$
Is my approach OK ?
$\displaystyle S_N(f)=\int_0^1 f(x-y)D_N(y)dy$, with $D_N(y)=\frac{\sin((2N+1)\pi y)}{\sin(\pi y)}$ and $\int_0^1D_n(y)dy=1$ (these identities are already proved). Then,
$\displaystyle |S_N(f)-f|=\int_0^1D_N(y)\left(f(x-y)-f(x)\right)dy=\int_0^1 \frac{f(x-y)-f(x)}{\sin(\pi y)}\sin((2N+1)\pi y) dy$
$\frac{f(x-y)-f(x)}{\sin(\pi y)}$ is OK in $0$ and $1$ by l'Hospital and integrable
Can I apply now Riemann-Lebesgue lemma ?
Your approach is correct. Under the given assumptions, the function $$g(y) = \frac{f(x-y)-f(x)}{\sin(\pi y)} $$ is continuous and $1$-periodic, with $g(0)=-f'(x)/\pi$. In particular, it is integrable which is enough to apply the Riemann-Lebesgue lemma.
More generally, this argument works for Hölder continuous $f$: an estimate of the form $|f(x-y)-f(x)|\le C|y|^\alpha$ with $\alpha\in (0,1]$ implies that $|g(y)| = O(|y|^{\alpha-1})$ as $y\to 0$, hence $g$ is integrable.