 Solution

The Arzelà-Ascoli theorem for $C([0,1])$ states that a family has compact closure in the uniform norm exactly when it is uniformly bounded and equicontinuous. To prove sufficiency, choose a countable dense set. From any sequence of functions, successive subsequence extraction at each dense point and a diagonal selection give convergence at all those points. Given $\epsilon >0$, equicontinuity gives a common $\delta$; cover the interval by finitely many $\delta$-balls centered at dense points. On a sufficiently late tail the functions differ by less than $\epsilon /3$ at those finitely many centres, and equicontinuity makes each of the other two differences less than $\epsilon /3$. Thus the selected subsequence is uniformly Cauchy. Completeness of $C([0,1])$ gives a continuous uniform limit. In a metric space this proves compactness of the closure. Conversely, a compact family has finite uniform $\epsilon$-nets. The finitely many net functions are bounded and uniformly continuous; approximation by them gives a common bound and equicontinuity for the family. This proves both directions.

For $f\in S_N$, a positive interior maximum would have $f^{\prime }=0$ and $f^{\prime \prime }\le 0$, contradicting $f^{\prime \prime }=f>0$. A negative interior minimum likewise contradicts $f^{\prime \prime }=f<0$. The endpoint bounds therefore give $|f|\le N$ everywhere.

Now put $w=f^{\prime }$. Differentiating the equation, using the stipulated third derivative, gives $w^{\prime \prime }=w+2w{w}^{\prime }$. At an interior extremum of $w$, $w^{\prime }=0$, so $w^{\prime \prime }=w$. The same maximum/minimum argument and the derivative endpoint bounds imply $|f^{\prime }|=|w|\le N$. Consequently $|f(x)-f(y)|\le N|x-y|$ by the mean value theorem. Thus $S_N$ is uniformly bounded and equicontinuous, and the theorem makes it relatively compact, hence totally bounded in $C([0,1])$. Its closure need not be asserted to remain thrice differentiable.