Period three implies all periods

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

The covering words $IJJJJ$ and $IJIJJ$ both occur under the interval coverings in period three implies all periods. Their points cannot be the boundary three-cycle. Each has least period five, because five is prime and each itinerary visits both intervals. Their respective one and two visits to the interior of $I$ distinguish the orbits even after cyclic changes of starting point.