Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence 
Let $f_n:[a.b]\rightarrow \mathbb{R}$ be sequence of $L-$Lipschitz functions, that is: $$\forall x,y\in[a,b]: |f_n(x)-f_n(y)|\leq L|x-y|$$
  Suppose $f_n \rightarrow f$ pointwise, prove $f_n \rightrightarrows f$

I have all the parts of the puzzle for the proof, and I'm trying to put them all together, I'm using this in my answer.
I would appreciate is you could correct my proof, and if you have an alternative proof, I would be more then happy to see it.
My proof: 
Let $\epsilon>0.$
$f_n$ are uniformly continuous on $[a,b]:$
$\tag{1} \exists \delta>0\ \forall x,y\in[a,b]: |f_n(x)-f_n(y)|<\frac{\epsilon}{3}$
$f$ is also $L-$Lipschitz:
$\tag{2} \forall x,y\in[a,b]:|f(x)-f(y)|<L|x-y|=\frac{\epsilon}{3}$
Let us set a partition of $[a,b]$ such as Stephen Montgomery-Smith suggests:

Pick points $x_1,\dots,x_m \in [a,b]$ which are distance
  $\frac{\epsilon}{3}$ from each other.  
For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge
 N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$.
  Let $N = \max_{1 \le i \le m} N_i$
Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| <\frac{\epsilon}{3L}: 
|f_n(x)-f(x)|<\frac{\epsilon}{3} \tag{3}$

$$\begin{align}|f_n(y)-f(y)| &=|f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &\leq |f_n(y)-f_n(x)| + |f_n(x)-f(x)|+|f(x)-f(y)| \\ &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \\\end{align}$$
 A: Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{3}, \frac{\varepsilon}{3L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$ we may find a finite subcover, say $\{B(\,x_1, \delta),  \, \ldots, \, B(\,x_M, \delta)\}$ (Heine-Borel Theorem).  Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that 
\begin{equation} \left|\, f_n(x_j) -
 f_m(x_j) \right| <  \frac{\varepsilon}{3}  \text{ whenever } n, m \geq N_j \,.
\end{equation}
Setting $N = \max [N_1,  \ldots, N_M]$ shows that
\begin{aligned}
\left|\,f_n(x)- f_m(x) \right| & \leq  \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\
& < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon
\; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [a,b] . 
\end{aligned}
Since $\mathbb{R}$ is complete, it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[a,b]$ (Cauchy Criterion).
I prefer uniform convergence first, ask $\,f$ questions later -_-.
