Closed subset of $C([0,1],\mathbb R$) 
Let $B=\{f\in C^1[0,1]:\Vert f\Vert _\infty \le A\}$. Is $B$ a closed subset of $C([0,1],\mathbb R)$?

Here's what I tried to do:
Let $\{f_n\}$ be a sequences of functions of $B$ such that $f_n \to f$, with $f\in C([0,1],\mathbb R)$.
If I prove that  $f\in B$, then I finish the proof, i.e. $B$ will be closed in  $C([0,1],\mathbb R)$.
But I don't know how could I do that.
Note:  $C([0,1],\mathbb R)$ is the space of continuous functions with domain $[0,1]$.

Take $A=2.$ Then the functions
$$f_n(x) = (1/n +(x-1/2)^2)^{1/2}$$
are in $B(2)$ and converge uniformly to $|x-1/2|$ on $[0,1].$ Since $|x-1/2|\not \in C^1,$ $B(2)$ is not closed.
