Let $S=\{f\in C[0,1]:\max\limits_{x\in[0,1]}(|f|+|f'|)\leq M\}$. Show that $S$ is compact in $C[0,1]$.
I have proved that any sequence in $S$ contains a convergent subsequence by Arzela Ascoli theorem,
but my question is: How can I check that the subsequence converges to the point in $S$ and thus prove that $S$ is compact(sequentially compact)?
Another way of approach this problem is the following. First prove that
$$||f||:=\max_{x\in [0,1]}(|f(x)|+|f'(x)|)$$
is a norm on $C^1[0,1]$ that is equivalent to the usual supremum norm $||f||_{\infty}=\max_{x\in[0,1]}|f(x)|$. This implies that the topologies induced by these norms are the same.
Now $S$ is precisely the closed ball $B(0,M)$ in the topology induced by $||\cdot||$, which is compact as it is closed and bounded (this is an equivalent to the sequential definition for compactness in metric spaces as you probably know).