Uniform convergence, Bounded derivative.

Let $f_n(x):[0,1] \to \Bbb R$ be a sequence of differentiable functions such that for every $n$ and every $x \in [0,1]$ we have $|f'_n(x)| \leq 1$. Also, $f_n \to f$ (pointwise convergence) in $[0,1]$. Prove that $f_n$ converges to $f$ uniformly.

I managed to prove that for every $x_0$, there is a uniform convergence in $(x_0-\delta, x_0 +\delta)$ for some $\delta > 0$.

How can I finish the proof?

Hint. Note that $$f_n(x)=f_n(0)+\int_{0}^xf_n'(t)\,dt$$ Then $$\sup_{x\in[0,1]}|f_n(x)-f_m(x)|\leq |f_n(0)-f_m(0)|+\int_{0}^1|f_n'(t)-f'_m(t)|\,dt$$ Now by using the pointwise convergence and dominated convergence theorem show that $(f_n)_n$ is a Cauchy sequence in $C([0,1])$ with the sup norm.

You can end your proof like this: since $[0,1]$ is compact, you can find $x_1,...,x_n$ and $a_1>0,...,a_n>0$ such that $f_n$ converges uniformly of $(x_i-a_i,x_i+a_i)$ and $\bigcup_i(x_i-a_i,x_i+a_i)\cap [0,1]=[0,1]$.

For every $c>0$, there exists $N_i$ such that for every $n>N_i, x\in (x_i-a_i,x_i+a_i)$ implies that $|f_n(x)-f(x)|<c$. Take $n=Sup(N_1,...,N_n)$.