Example for equicontinuous bounded but not compact in $C(K)$ The theorem states if $\mathcal{F}$ is closed in $C(K)$, where $K$ is compact in $\mathbb{R}^n$, then $\mathcal{F}$ is compact $\iff$ $\mathcal{F}$ is equicontinuous and bounded.
And it has a corollary saids that if $\{f_n\}$ is an equicontinuous bounded sequence, then it has a convergent subsequence in $C(K)$. Then corollary is proved by considering the closure of $\{f_n\}$ in $C(K)$ and apply Arzela-Ascoli theorem. 
I think the closeness in the theorem is necessary, since we can find a convergent equicontinuous and bounded sequence $\{f_n\}$ in $\mathcal{F}$ but by deleting the limit point of $\{f_n\}$ from $\mathcal{F}$ so that it has no convergent subsequence. But may you give me an explicit example?
Also I am confused about the following example:
$\mathcal{F}=\{f||f'|(x)\leq 1, f(0)=0\}$
It's easy to say it's equicontinuous and bounded, but it think it's hard to show it's closed. The suggested solution directly concludes it's compact hence closed. Why is it true? 

Let $f_n(x) = \frac1n$ for all $x\in K$.  Then $\mathcal F=\{f_n\}$ is bounded and equicontinuous but not closed.
No, the set $\mathcal{F}=\{f||f'|(x)\leq 1, f(0)=0\}$ is not closed in $C([-1,1])$ with the sup norm.  E.g., $\frac12|x|$ is a limit point not in the set.  The set of Lipschitz functions with Lipschitz constant at most one would be closed, though.
