= Solution
A subset is <totally bounded> if for every $\varepsilon>0$ it has a finite cover by balls of radius $\varepsilon$. If its <closure> is <compact>, the cover by all such balls has a finite subcover, proving total boundedness.
Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as $\sigma_r$. These sets are infinite and nested, and two terms indexed by $\sigma_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in\sigma_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the <closure> as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the <closure> has the same limit. <Sequential compactness> therefore proves \b[total boundedness implies relative compactness in a complete ambient space].
The <Arzelà-Ascoli theorem> says a subset of $C[0,1]$ is <relatively compact> exactly when it is uniformly bounded and <equicontinuous>. Pointwise boundedness together with <equicontinuity> is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\leq|t-s|$, uniformly over the family, while $\|f\|_\infty\leq1$. Thus this family is <relatively compact>.
For $V$, the image of the <unit ball> has sup <norm> at most one and derivative $f$ of sup <norm> at most one; it lies in that family, so \b[$V$ is <compact>].
For $W$, take $f_n=n\mathbf1_{[0,1/n]}$, each of $L^1$ <norm> one. Then $Wf_n(t)=\min(nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus \b[$W$ is not <compact>]. The same integration formula has different compactness properties because the domain unit balls impose different controls.
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