= Solution
Use <binary expansions> ending in zeros at dyadic rationals. The <doubling map> shifts the expansion left:
$$
x=0.b_1b_2b_3\ldots\quad\Longrightarrow\quad F(x)=0.b_2b_3b_4\ldots.
$$
In ordinary coordinates it is $2x$ on $[0,1/2)$ and $2x-1$ on $[1/2,1)$. The branches are continuous; the left limit at $1/2$ is $1$ while its value and right limit are $0$. This is its only discontinuity in the half-open interval.
For every $N\ge2$, the repeating binary block $0^{N-1}1$ gives
$$
\boxed{x_N=\frac1{2^N-1},\qquad F^N(x_N)=x_N.}
$$
Its least period is $N$: a shorter repeating block would make more than one digit equal to one in a length-$N$ period. The orbit avoids the discontinuity, and its multiplier is $(F^N)'=2^N>1$, so \b[these periodic orbits are repelling and unstable]. The same multiplier argument applies to every periodic orbit, with the endpoint <fixed point> treated one-sidedly.
In the two-branch horseshoe formulation of <Glendinning chaos>, disjoint subintervals are each mapped onto the same interval by an iterate. Here $J=(0,1)$, $K_0=(0,1/2)$ and $K_1=(1/2,1)$ already satisfy $F(K_0)=F(K_1)=J$. Their inverse branches $x\mapsto x/2$ and $x\mapsto(x+1)/2$ realize every binary itinerary. This is a horseshoe for a piecewise continuous map; on the circle the endpoint jump disappears. The symbolic explanation also shows <dense periodic points>, by repeating any prescribed finite prefix, and <topological transitivity>, because an iterate maps any sufficiently small dyadic cylinder onto $(0,1)$. Finally every neighborhood contains points sharing an arbitrarily long prefix but with different subsequent digits; an appropriate iterate separates their images by a fixed amount, for instance more than $1/4$. This proves <sensitive dependence on initial conditions> as well. \b[The binary shift has full two-symbol chaotic dynamics.]
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