If $f(0) = 0$ and $|f'(x)| \leq 1$ $\forall x \in(0,1)$. Prove that $|f(x)| \leq 1$ $\forall x\in(0,1)$.
I tried to solve it via definition of continuity $(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in A)(|x-a|<\delta\implies |f(x)-f(a)|<\varepsilon)$ for $a = 0$ and $\delta = 1$ and I got $|x|<1 \implies |f(x)|<\varepsilon$ but I don't know how to find $\varepsilon$ and how to use derivative to get the solution. Please help.