$A$ be the closure in $\mathcal{C} [0,1]$ of the set $B$ where $B = \{f \in \mathcal{C}^1 [0,1] \mid \mid f(x) \mid \leq 1 \text{ and } \mid f'(x) \mid \leq 1 \text{ for all } x \in [0,1] \}$, then $B$ is
(Multiple correct)
a) closed
b) compact
c) connected
d) dense
My approach: It is obvious that is closed because we our considering the closure. For compactness I thought of applying Ascoli theorem but I don't know how to show bounded and equicontinious. Also I don't know how to show connected and dense
Since the derivatives are uniformly bounded, for any $f \in B$ we have $$|f(x)-f(y)| \le \max_{z\in [0,1]}\{f'(z)\}|x-y| \le |x-y|, \quad \forall x,y \in [0,1].$$ by the mean-value theorem, so $B$ is equicontinuous. This together with uniform boundedness of $B$ means Arzela Asocoli gives us the result.
See @freakish's comments for the others.