Show that $f(x) = 0$ for all $x \in [a,b]$ given $|f'(x)| \leq C|f(x)| $

Assume that there exists $x_0$ such that $f(x_0)\ne 0.$ Since $f$ is continuous there exists a maximal interval $(c,d)$ such that $f(x)\ne 0$ on $(c,d)$ and $f(c)=0.$ (Note that it can be $c=a.$)

Now we have that $g(x)=\ln |f(x)|$ satisfies $|g'(x)|\le C.$ Thus, for any $t\in (0,1)$ and $\epsilon=x_0-c$

$$|g(x_0)-g(c+t\epsilon)|=\left|\int_{c+t\epsilon}^{x_0}g'(x)dx\right|\le C(x_0-c-t\epsilon).$$ Take the limit as $t\to 0$ to get a contradiction.