Prove that $f(x) = 0$ for all $x \in [a,b]$, where $f$ is differentiable on $[a,b],~f(a) = 0$ and $|f'(x)| \leq A|f(x)|$ on $[a,b]$ for some $A\geq0$

Q. Suppose $f$ is differentiable on $[a,b]$, $f(a) = 0$ and there is real number $A$ such that $|f'(x)| \leq A|f(x)|$ on $[a,b]$. Prove that $f(x) = 0$ for all $x \in [a,b]$.

My attempt: if $A=0$, it is clear that $f(x)=f(a)=0$ for all $x \in [a,b]$.

If $A>0$, take $x_0=a+\frac{1}{nA},~n \neq 0$, then, from the mean value theorem and $M_1 \leq AM_0$, it is clear that $$|f(x)| \leq M_1 (x - a)~\&~M_1(x-a) \leq A M_0 (x_0 - a)$$for any $x \in [a,x_0]$, where $M_1=\sup_{[a,x_0]}|f'(x)|$ and $M_0=\sup_{[a,x_0]}|f(x)|$, this gives $$|f(x)| \leq A M_0 (x_0 - a)=\frac{M_0}{n},$$for any $x \in [a,x_0]$. Hence $$M_0=\sup_{[a,x_0]}|f(x)| \leq \frac{M_0}{n},$$

Now,

(i) How can I ensure that $n>1$ to obtain $M_0=0$ ?

(ii) How the idea is extended to any $x_0 \in (a,b]$ ?