Is the sequence of continuous functions such that $\left\Vert f_{n}(x) -f_{m}(x) \right\Vert _{\infty}=1$ equicontinuous? 
Consider a bounded sequence of continuous functions $f_n:\left[0,1\right]\to\mathbb{R}$ such that $$\left\Vert f_{n}-f_{m}\right\Vert _{\infty}=\sup_{x\in\left[0,1\right]}\left|f_{n}\left(x\right)-f_{m}\left(x\right)\right|=1$$ whenever $n\neq m$. Can such a sequence be equicontinuous?

I want to say no and my reasoning is that if you have such a sequence, then it must act like $\left\{x^n\right\}$ which can be shown to be not equicontinuous. However, I am not sure.
 A: Note, as Jose27 points out in the comments, that the boundedness assumption is not needed, as it is implied by the norm condition.
From the norm condition, 
$$\tag{1}
\Vert f_n-f_m\Vert_\infty=1,\quad \text{whenever }\ n\ne m,
$$
it follows that
the sequence $\{f_n\}_{n=1}^\infty$ is uniformly bounded over $[0,1]$. 
Equation (1) also implies that $\{f_n\}_{n=1}^\infty$ cannot be equicontinuous  over $[0,1]$:
Suppose $\{f_n\}_{n=1}^\infty$ were  equicontinuous  over $[0,1]$. Then by the 
Arzelà-Ascoli Theorem, there is a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}_{n=1}^\infty$ that converges uniformly on $[0,1]$.   In particular, $\{f_{n_k}\}_{k=1}^\infty$ is uniformly Cauchy.  That is, for each $\epsilon>0$, there is a positive integer $N$ so that
$$
|f_{n_k}(x ) -f_{n_{l}}(x )|<\epsilon, \quad \text{for all }\ k,l\ge N\ \text{ and all }\ x\in[0,1].
$$
But, setting $\epsilon={1\over2}$ and fixing a positive integer $N$, we have by equation (1) the existence of some $x_{\scriptscriptstyle N}\in[0,1]$ with
$|f_{n_N}(x_{\scriptscriptstyle N}) -f_{n_{N+1}}(x_{\scriptscriptstyle N})|>{1\over2}$. As $N$ was arbitrary, this contradicts the fact that  $\{f_{n_k}\}_{k=1}^\infty$ is uniformly Cauchy.
It follows that  $\{f_n\}_{n=1}^\infty$ is not  equicontinuous  over $[0,1]$. 
