 Solution

Fix $x\in (0,1)$ and put $x_0=x$, $x_{n+1}=f(x_n)$. Then $0<x_{n+1}<x_n\le x_0<1$. The bounded monotone sequence theorem gives a limit $\ell \in [0,x_0]$. If $\ell >0$, it lies in the domain, where continuity allows passage to the limit in the recurrence: $\ell =f(\ell )$. That is a fixed point, contrary to $f(\ell )<\ell$. Hence $\boxed{f^n(x)\to 0\text{ for every }x\in (0,1)}$. Continuity is used at a possible positive limiting point, not at the excluded endpoint zero.