I'd like to prove that for a sequence of differentiables functions $(f_n)$ on $[0;1]$ which converges to $0$, actually converges uniformly.
We only know that $\vert f'_n(x)\vert \leq 2015 + \cos(x) $ for all $x$ and $n$.
I know that it means that the derivatives of all functions from the sequence is bounded by 2016 and thus they are lipschitz, but I don't know if it helps to prove that the convergence is uniform.
Any help would be welcome.
You have indeed identified the key element of the proof. The other is the compactness of $[0, 1]$.
If you don't want to work it out yourself :
Let $\varepsilon > 0$. Since $[0, 1]$ is compact, there exists an integer $N$ and $x_1, \ldots, x_N \in [0, 1]$ such that $$[0, 1] \subset \bigcup_{0 \leq i \leq N} B(x_i, \frac{\varepsilon}{2016}).$$ (no need for big guns here, you can just take $N = \mathrm{floor}(2016/\varepsilon)$ and $x_i = i/N $. Since $f_n(x_i) \underset{n \to \infty}{\longrightarrow} 0$ for every $0\leq i \leq N$, there exists $n_0 \geqslant 0$ such that $$\forall n \geq n_0, \forall 0 \leq i \leq N, |f_n(x_i)| \leq \varepsilon. $$ If $y \in [0, 1]$, there exists an index $i_0$ such that $x \in B(x_{i_0}, \frac{\varepsilon}{2016})$. Thus, applying the triangle inequality, $$ \forall n \geq n_0, \quad |f_n(y)| \leq |f_n(y) - f_n(x_{i_0})| + |f_n(x_{i_0})| \leq 2016\cdot |y - x_{i_0}| + |f_n(x_{i_0})| \leq 2 \varepsilon. $$ Hence, $$\forall n \geq n_0, \|f_n\|_{\infty} \leq 2\varepsilon.$$ The convergence is uniform.