How Can I apply Arzela-Ascoli to show that $f_n \to f$ uniformly? I have that $M >0$ and $f_n:[0,1] \to R$ is a sequence of Lipschitz functions with common Lipschitz constant $M$, i.e. $|f_n(x)-f_n(y)| \le M|x-y|$ and I have that $f_n \to f$ pointwise. 
How can I apply Arzela Ascoli theorem to show that $f_n \to f$
Since $f_n \to f$ pointwise, I can show that $f$ is Lipschitz continuous by doing the following 
$|f_n(x)-f_n(y)| \le M|x-y| \quad \forall n $ and $x,y \in [0,1]$ 
Since $f_n(x) \to f(x)$ and $f_n(y) \to f(y)$ when $n \to \infty$, it follows that 
$|f_n(x) - f_n(y)| \to |f(x)-f(y)|$ and from $|f_n(x)-f_n(y)| \le M|x-y|$ we get that 
$|f_n(x)-f_n(y)|\le M|x-y|$. 
Now I dont know how to show that it follows that $f_n \to f$ uniformly and the hint is to use Arzela-Ascoli theorem. Can somebody help me?
 A: Theorem 1. (Arzela Ascoli) Consider a sequence of real-valued continuous functions $\{ f_n \}_{n \in \mathbb N}$ defined on a closed and bounded interval $[a, b]$ of the real line. If this sequence is uniformly equicontinuous and uniformly bounded. Then there exists a subsequence $\{ f_{n_k} \}_{k \in \mathbb N}$ that converges uniformly.
Theorem 2. (Subsequence Principal)A sequence $\{ x_n\}$ in topology space $X$ converges to $x^*$ iff every its subsequence has a further subsequence (sub-sub-sequence) converging to $x^*$.
Step 1. We prove that: $\{ f_n \}_{n \in \mathbb N}$ is uniformly equicontinuous and uniformly bounded. Indeed, the uniformly equicontinuous holds  directly thanks to the "common Lipchitz" assumption. Regarding the uniform boundedness, observe that $|f_n(x)|\leq |f_n(0)|+M$ and $|f_n(0)|$ is bounded since $f_n(0)$ converges to $f(0)$. Applying Theorem 1, we conclude that there exists a subsequence $\{ f_{n_k} \}_{k \in \mathbb N}$ that converges uniformly to a function, say $h$.
Step 2. As $f_n$ converges to $f$ pointwise and $f_{n_k}$ also converges to $h$ poitwise, then $h=f$. Thus, $\{ f_{n_k} \}_{k \in \mathbb N}$ uniformly converges to $f$.
Step 3. Repeat the arguments in Step 1 and Step 2 for arbitrary subsequence of $\{ f_{n} \}_{n \in \mathbb N}$, we will see that this subsequence has a further subsequence uniformly converges to $f$. Applying the Theorem 2, we conclude that $f_n$ converges to $f$ uniformly.
