**Edited:** As pointed out in comments, previous proof was all wrong, should now be fixed!
Let $x_0$ be the starting point, $x_1=f(x_0)$, $x_2=f(x_1)$, etc. We will show that if $x_n\in P_1$, there will be some later $x_{n+k}\in P_0$. This means there can't be an infinite tail of 1's, since every 1 is eventually followed by a 0.
If $x_n\in P_1$, write $x_n=1-\epsilon$, with $0<\epsilon\le \frac12$. We want to find a $k$ where $$ x_{n+k}=2^kx_n=2^k(1-\epsilon)=2^{k}-2^k\epsilon\in P_0+\mathbb{N} $$ where $$P_0+\mathbb{N}=[0,1/2)\cup[1,3/2)\cup[2,5/2)\cup\dots\dots$$ Since $2^{k}$ is an integer, it suffices to show that $2^k\epsilon\in (1/2,1]$. Letting $k$ be the smallest integer for which $2^k\epsilon>\frac12$, we must have $2^{k-1}\epsilon \le\frac12$, so that $2^k\epsilon \le 1$. This implies $2^k\epsilon\in (1/2,1]$, so that $x_{n+k}\in P_0$.