Prove if $f'(x)\geq 1$ then $\exists c$ such that $f(c)=0$.

Let $y=f(x) > 0$. By mean value theorem, we then have that there exists $c \in (x-y,x)$ such that $$f'(c) = \dfrac{f(x)-f(x-y)}{x-(x-y)} \geq 1 \implies f(x) - f(x-y) \geq y \implies f(x-y) \leq f(x)-y = 0$$ Hence, by intermediate value theorem, there exists $z \in [x-y,x]$ such that $f(z) = 0$.

Argue similarly for $y = f(x) < 0$.