TITLE: "The closure of the unit ball of $C^1[0, 1]$ in $C[0, 1]$" and its compactness
QUESTION [12 upvotes]: [I really want to apologize if this problem looks a little too long.]
The problem :
This is taken from here [Question. 3.7 (c)] and it says...

Prove or disprove the comapctness of the closure of the unit ball of $C^1[0, 1]$ in $C[0, 1]$.

What have I tried?
I think it's pretty clear that the unit ball in $C^1[0,1]$ is $B= \{f \in C^{1}[0,1]: \lVert f \rVert \leq 1\}$. Let us denote $K= \operatorname{cl}_{C[0,1]}(B)$, the closure of $B$ in $C[0,1]$.
I really didn't have much clue how to start working on this problem but finally thought about using the sequential approach. If $a$ and $b$ are two points such that $0<a<b<1$ then I think there can be functions $f_a$ and $f_b$ having the property that $\lVert f_a-f_b \rVert = 1$ which are sufficiently smooth to be in $C^1$.  Now as it is quite tedious to construct these functions explicitly, I scanned a hand-drawn picture of what I think they might look like. 

Here $f_a$ and $f_b$ take zero values in almost all of the interval $[0,1]$ and jumps up at $a$ and $b$ respectively. Also, as is clear from the figure, they don't assume non-zero values simultaneously. Now, my argument is clear: however close the two points $a$ and $b$ come to each other (remaining distinct) there will always be functions like $f_a$ and $f_b$. They may get steeper and steeper but will never lose their $C^1$-ness. 
So if I consider the sequence $(\frac{1}{n})$ in $[0,1]$, I will get a sequence $(f_{\frac{1}{n}})$ in $B (\subseteq K)$ where for $m \neq n$ we will have $\lVert f_{\frac{1}{m}} - f_{\frac{1}{n}}\rVert = 1$. Hence the sequence $(f_{\frac{1}{n}})$ cannot have a convergent subsequence, thereby proving that $K$ is not sequentally compact and hence not compact.
But I'm not very sure about all these. A friend of mine told me that this same problem has a positive answer and that made me sufficiently confused.
So, here comes my question...

In the above argument, where have I gone wrong?
  What is the real answer? And how to prove it?

Thanks a million for reading my extra-long question. And thanks a zillion for any help that you can offer.

REPLY [4 votes]: Ascoli-Arzelá requires says that a set $\scr F$ in $C^0([0,1])$ is compact iff it is:
1) closed, 2) bounded (or ``equibounded'') and 3) equicontinuous. 
If if $L=\{f\in \cal C^1:\max\{||f'||_\infty,  ||f||_\infty\} <1 \}$ is considered as a subset of $C^0$ then 
1) it is equibounded because it is a subset of the unit ball in $\cal C^0$ (i.e. $||f||_\infty<1$);
2) it is equicontinuous because all $f\in L$ are Lipshitz with constant at worst $1$ (by the "mean value theorem" of calculus).
3) As written it is clearly not closed in $\cal C^0$; but the question asks for its closure in $\cal  C^0$, so that comes for free.
What I am not sure of, is whether the closure in $\cal C^0$ coincides with the closure in $\cal C^1$, but I doubt it.