 Solution

For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.