Let $M$ be a bounded subset of $C_{[a, b]}$. Prove that the set of all functions
$$F(x) = \int_{a}^{x} f(t) dt$$
with $f \in M$ is compact.
I have the following results which might be helpful
Every countable compact metric space is totally bounded.
A metric space is compact if and only if it is totally bounded and complete.
I thought about showing that $M$ is totally bounded but couldn't do it.
This is not so.
Just to give it a name, let $S$ be the set of all $F$ such that $F(x)=\int_a^x f(t)\,dt$ for some $f\in M$. Assuming $M$ bounded does not imply $S$ is compact; it does imply that $S$ is precompact.
Say $[a,b]=[-1,1]$. Let $M$ be the set of all continuous $f$ with $|f(t)|\le1$ for all $t$. Let $$f_n(t)=\begin{cases}0,&(t\le0), \\nt,&(0\le t\le 1/n), \\1,&(t>1/n),\end{cases}$$and define $F_n(x)=\int_{-1}^x f_n$. Then $F_n\to F$ uniformly, where $$F(t)=\begin{cases}0,&(t\le 0),\\t,&(t>0).\end{cases}$$Then $F_n\in S$, $F_n\to F$ uniformly but $F\notin S$; hence the sequence $(F_n)$ cannot have a subsequence converging uniformly to ann element of $S$.