Let us take $f'(x) = 3a_{0}x^2 + 2a_{1}x + a_{2} = x^2 +\frac{2a_{1}}{3a_{0}}x + \frac{a_{2}}{3a_{0}} = 0$ (i). If $|f(x)| \le 1$ for all $x \in [-1, 1]$, then let $x = -1, 1$ be the roots of (i) such that $f(\pm 1) = \mp 1$ (when $a_{0} > 0$) or $f(\pm 1)=\pm 1$ (when $a_{0} <0$). This yields:
$x^2 +\frac{2a_{1}}{3a_{0}}x + \frac{a_{2}}{3a_{0}} = (x-1)(x+1) = 0 \Rightarrow \frac{2a_{1}}{3a_{0}} = 0, \frac{a_{2}}{3a_{0}} = -1$;
or $a_{1} = 0, a_{2} = -3a_{0}$, which produces the cubic $f(x) = a_{0}x^3 - 3a_{0}x + a_{3}$ (ii). If we now take $f(-1)=1, f(1)=-1$, we then obtain the system of equations:
$-a_{0} +3a_{0} + a_{3}=1$ (iii);
$a_{0}-3a_{0}+a_{3}=-1$ (iv)
and adding (iii) to (iv) yields $a_{3} = 0 \Rightarrow a_{0}=\frac{1}{2}$. If we also check $f(-1)=-1, f(1)=1$, we have:
$-a_{0} +3a_{0} + a_{3}=-1$ (v);
$a_{0}-3a_{0}+a_{3}=1$ (vi)
and adding (v) to (vi) yields $a_{3} = 0 \Rightarrow a_{0}=-\frac{1}{2}$.
Thus, $\exists c (c = 0)$ that is the same constant for all cubic polynomials with $|a_{0}| \le 4$ and satisfying the set of properties above.
QED