Prove that $f(x)\equiv0$ on $\left[0,1\right]$ if $f(0)=0$ and $|f'(x)|\le|f(x)|$

You can use the fact that $|f'(c)| \leq |f(c)|$ and since $f$ is differentiable it is also continuoues, therefore you could get the equality $|f(x)| = |\frac{f(x)} x |$ which means $f(x)=0$

Let $x_0 \in [0,1]$ with $f(x_0) = 0$. We will show that $f(x) = 0$ for all $x \in [x_0, x_0 + \varepsilon]$ for some $\varepsilon > 0$ (if $x_0 < 1$). This is sufficient to prove your claim because you can then consider $\gamma := \sup \{x \in [0,1] \mid f|_{[0,x]} \equiv 0\}$, show $\gamma = 1$ and deduce $f \equiv 0$.

Because $f$ is continuous and $f(x_0) = 0$, there is some $\varepsilon_1 \in (0, \frac{1}{2})$ with $|f(x)| \leq \frac{1}{2}$ for all $x \in [x_0 , x_0 + \varepsilon]$.

We will now show inductively that actually $|f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. For $n=1$ this follows by our choice of $\varepsilon$.

For the inductive step note that $|f'(x)| \leq |f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. Using the mean-value theorem, it is now easy to see that

$$|f(x)| = |f(x) - f(x_0)| = |f'(\xi)| \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^n \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^{n+1}$$

for $x \in [x_0, x_0 + \varepsilon]$, because $\varepsilon \leq \frac{1}{2}$.

In the limit $n\rightarrow \infty$, we get our claim.

$\displaystyle \left |\frac{f(x)}x \right | \le |f(c)| \implies |f(x)| \le |f(c)| $ where $0<c<x$. This implies $\displaystyle |f(x)| \le \left| \lim_{\epsilon \to 0^+}f(\epsilon)\right| = 0$

To show that $c_n \to 0$, either there exists finitely $c \in (0, x)$ such that $f'(c) = f(x)/x$ or there and interval $(a,b)$ where $f'(c) = \mathrm{constant}$, second case, $c = \inf\{(a,b)\}$.