= Solution
In <Glendinning chaos>, a continuous <interval map> is chaotic when some positive iterate has a <horseshoe for an interval map>. In the interval version, one may take an open interval $J$ containing two disjoint open intervals $K_0,K_1$ with $F^m(K_0)=F^m(K_1)=J$.
Let a three-cycle be ordered $a<b<c$. There are only two possible cyclic orderings; reflection of the coordinate interchanges them, so assume $F(a)=b$, $F(b)=c$, $F(c)=a$. Then
$$
F^2(a)=c,\qquad F^2(b)=a.
$$
Also $F(b)=c$ and $F(c)=a$, so the <intermediate value theorem> gives $d\in(b,c)$ with $F(d)=b$, whence $F^2(d)=c$. Both $[a,b]$ and $[b,d]$ therefore map across $[a,c]$ under $F^2$.
To obtain exact open-interval images, use the following elementary first-passage observation. A continuous function on $[r,s]$ with endpoint values $a,c$ has a subinterval on which its endpoint values are $a,c$ and all interior values lie strictly between them: take the first hit of the second level, then the last hit of the first level before it. The <intermediate value theorem> makes its interior image exactly $(a,c)$. Apply this to the two intervals, reversing endpoint order on the first. Their selected interiors $K_0\subset(a,b)$ and $K_1\subset(b,d)$ are disjoint and lie in $J=(a,c)$, with
$$
\boxed{F^2(K_0)=F^2(K_1)=J.}
$$
This proves the <three-cycle forces a two-iterate interval horseshoe> result and establishes the required horseshoe for $F^2$.
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