= Solution
The space $\Sigma$ of all infinite <binary sequences> is <uncountable> by the <Cantor diagonal argument>: given any list $s^{(1)},s^{(2)},\ldots$, the <sequence> $t_i=1-s_i^{(i)}$ differs from its $i$th entry at position $i$, so it is absent from the list. Part (i) shows that only countably many <sequences> are periodic. Their complement is therefore <uncountable>.
Each <forward orbit> $\{\sigma^n(s):n\geq0\}$ is countable. If there were only countably many nonperiodic <forward orbits>, their union together with the periodic points would be countable, contradicting the preceding argument. Hence \b[there are uncountably many nonperiodic <orbits>], rather than merely uncountably many nonperiodic starting points.
One can also exclude eventual periodicity. An eventually periodic <sequence> is specified by a finite initial word and a finite repeating word, so only countably many exist. Removing all of them still leaves an <uncountable set>. This proves the assertion even if “nonperiodic” is understood to exclude <orbits> that eventually enter a <periodic orbit>.
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