\begin{align} |f(x)| &= |f(x) - f(a)| = |f'(c)||x-a| \le A|f(c)||x-a| \\ & = A|f(c) - f(a)||x-a| = A|f'(d)||d-a||x-a| \le A|f'(d)||x-a|^2 \\ & \le |f(d)||A(x-a)|^2. \end{align} Now suppose there exists $c \in ]a,x[$ such that $|f(x)| \le |f(d)| |A(x-a)|^n$, by induction on $n$. Then \begin{align} |f(x)| & \le |f(c)||A(x-a)|^n = |f(c) - f(a)||A(x-a)|^n \\ & = |f'(d)||d-a||A(x-a)|^n \le A|f(d)||d-a||A(x-a)|^n \\ & \le |f(d)||A(x-a)|^{n+1}. \end{align} Hence for every integer $n$ there exists $c_n$ such that $|f(x)| \le |f(c_n)||A(x-a)|^{n+1}$. Since $[a,b]$ is compact and $f$ is differentiable, $f$ is continuous, hence $|f|$ is continuous and attains a maximum $M$ over $[a,b]$, so that $|f(x)| \le |f(c_n)| |A(x-a)|^n \le M|A(x-a)|^n$. This means $f(x) = 0$ as long as $|x-a| < \frac 1A$, since the bound only depends on $n$, hence we can let it go to infinity. By continuity of $f$ you also get $f(x) = 0$ if $|x-a| \le \frac 1A$.
Since you can split the interval $[a,b]$ as $[a,a+\frac 1A]$, $[a+\frac 1A, a + \frac2A]$, $\dots$, $[a+\frac mA, b]$, you can repeat this process finitely many times and show that $f$ is zero over $[a,b]$.
Hope that helps,