We define the norm on $$\mathcal C^1(\Bbb T)$$ as $$\|f\| := \|f\|_{\mathcal C(T)} + \|f'\|_{\mathcal C(T)}.$$ With this norm, $\mathcal C^1(\Bbb T)$ is a Banach space. I'd like to show the existence of a function $f \in \mathcal C^1(\Bbb T)$ whose Fourier series $\{S_N(f)\}_{N\ge 0}$ does not converge to $f$ in the norm of $\mathcal C^1(\Bbb T)$.
We know that there exists a continuous function $g\in C(\Bbb T)$ such that the partial sums $S_N g$ of its Fourier series diverge at $0$. Consider $f(x) := \int_0^x g(t)\, dt$ for $0\le x \le 1$. We can translate $g$ appropriately (without affecting continuity and divergence of the Fourier series at $0$) to ensure $\int_0^1 g(t)\, dt = 0$, so that $f \in C(\Bbb T)$. In fact, $f\in C^1(\Bbb T)$, as $f'(x) = g(x)$. Will this $f$ do the trick?
Following @SeverinSchraven's comment, consider $g \in C(\Bbb T)$ such that $\int_0^1 g(t)\, dt = 0$ and $S_N (g)$ diverges at $0$. Define $f(x) := \int_0^x g(t)\, dt$. Then, $f\in C^1(\Bbb T)$. Clearly, $$\Vert f-S_N(f)\Vert_{\mathcal{C}^1(\mathbb{T})}\geq \Vert f'-S_N(f)'\Vert_{\mathcal{C}(\mathbb{T})}=\Vert g-S_N(g)\Vert_{\mathcal{C}(\mathbb{T})}$$ so that $S_Nf$ doesn't converge to $f$ in the norm of $C^1(\Bbb T)$ (if it did, then $S_N g \to g$ uniformly, which contradicts the construction of $g$.)