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 $[0,1]$.