Show that max $|f(x)| ≥ (c_0/8) (b − a)^2$

You are on the right track. Note, however, that $|f(x_0)| = M$, not simply $f(x_0) = M$. So the equation $f(x) = M + \frac{(x - x_0)^2}{2} f''(t)$ should be $f(x) = f(x_0) + \frac{(x - x_0)^2}{2} f''(t)$.

Setting $x = a$ yields $$f(x_0) = -\frac{(x_0 - a)^2}{2}f''(t_1)$$ for some $t_1$ between $x_0$ and $a$. Similarly $f(x_0) = -\frac{(b - x_0)^2}{2} f''(t_2)$ for some $t_1$ between $x_0$ and $b$. Thus $$|f(x_0)| \ge \frac{c_0}{2}(x_0 - a)^2\quad \text{and}\quad |f(x_0)| \ge \frac{c_0}{2}(b - x_0)^2$$ Therefore, $$M\ge \frac{c_0}{2}\left[\frac{(x_0 - a)^2}{2} + \frac{(b - x_0)^2}{2}\right] \ge \frac{c_0}{2}\left[\frac{(x_0 - a) + (b - x_0)}{2}\right]^2 = \frac{c_0}{8}(b-a)^2$$ as desired.