Is the following set a compact set? Let $A$ be defined as
$$A:=\{f\in C^1([0,1],\mathbb{R}) : \|f\|_{C^1} \leq 1\}.$$
I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set? 
 A: The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $\lVert\cdot\rVert_{C^1}$.
But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.
I don't know which norm the norm $\lVert\cdot\rVert_{C^1}$ is, but I suppose that it is such that $C^1\bigl([0,1],\mathbb R\bigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $\iff$ precompact and complete.
A: The answer is NO.
Consider the sequence
$$
f_n(x)=\frac{1}{n+1}\sin(nx),\,\,n\in\mathbb N.
$$
Then 
$$
\|\,f_n\|_{C^1}= \max \|\,f_n\|+ \max \|\,f_n'\|=\frac{1}{n+1}+\frac{n}{n+1}=1.
$$
If $\{\,f_n\}$ possessed a converging subsequence $\,\{\,f_{n_k}\}$, in the $C^1-$sense, with limit $\,f,\,$ then $\,\{\,f_{n_k}\}$ would also converge to $f$ in the uniform sense. But, $\{\,f_{n_k}\}$
converges uniformly to $f\equiv 0$. Nevertheless, $\,\{\,f_{n_k}\}$ DOES NOT converge in the $C^1-$sense to $0$, since $\,\|\,f_{n_k}\|_{C^1}=1$.  
