Supose $f'(x) \ge M \gt 0$ for every $x \in [0,1]$... Supose $f'(x)\ge M\gt 0$ for every  $x \in [0,1]$. Prove that there exists an interval of length $\frac 14$ where $$| f(x) |\ge \frac M4$$
 A: Hint. 
If $f(x_0)\ge 0$ for some $x_0\in[0,\frac12]$ then $f(x)\ge \frac M4$ for all $x\in[x_0+\frac14,x_0+\frac12]\subset[0,1]$. 
If $f(x_0)\le 0$ for some $x_0\in[\frac12,1]$ then $f(x)\le -\frac M4$ for all $x\in[x_0-\frac12,x_0-\frac14]\subset[0,1]$. 
There are no other cases to consider. 
Let me list the cases in a different way, trying to match your approach in the comments. 
Case (A). $f(x)>0$ for all $x$. In particular $f(0)>0$. You have done this case in a comment (with $f(0)\ge0$). 
Case (B). $f(x)<0$ for all $x$. In particular $f(1)<0$. Similarly to what you did, $f(x)\le-\frac M4$ on $[\frac12,\frac34]$ (notice you only need an interval of length $\frac14$, not $\frac12$ even though you found one of length $\frac12$ in the comments). 
Case (C). $f(x_0)=0$ for some $x_0$. If $x_0\le\frac12$ consider the interval $[x_0+\frac14,x_0+\frac12]$ where $f(x)\ge\frac M4$. If $x_0\ge\frac12$ consider the interval $[x_0-\frac12,x_0-\frac14]$ where $f(x)\le-\frac M4$.  
A: Suppose without loss of generality (why?) that $f(x) < 0$ or $f(x) > 0$ for all $x \in [z, z + \frac{1}{4}] \subset [0,1]$. Then, we have
$|f(z)| = |f(z)- f(z + \frac{1}{4})| + |f(z + \frac{1}{4})| \geq \frac{M}{4}$. (By the mean value theorem).
Therefore, $|f(z)| \geq \frac{M}{4}$. Since $|f'(x)| = f'_{+}(x) - f'_{-}(x)$. We have,
$|f(x)| \geq \frac{M}{4}, \forall x \in [z, z + \frac{1}{4}]$
Or
$|f(x)| \geq \frac{M}{4}, \forall x \in [z - \frac{1}{4}, z]$.
