Why do all non-empty ω-regular languages have periodic members?
Actually, it doesn't say that $q$ is an accepting state, though it should do!  $\mathrm{Inf}(r)$ is the set of states visited infinitely often during the run&nbsp;$r$.  The acceptance condition of B&uuml;chi automata is that $\mathrm{Inf}(r)$ contains an accepting state.

The correct proof is as follows, with explanation added. Note also that, while your title says that the automaton is deterministic, the proof applies to nondeterministic automata, too, since it talks about the existence of a run with a particular property, rather than saying that the unique run has that property.

Let $A$ be a B&uuml;chi automaton and let $s\in L^\omega(A)$. Since $A$ accepts&nbsp;$s$, there must be a run&nbsp;$r$ of&nbsp;$a$ on&nbsp;$s$ that has an accepting state $q\in \mathrm{Inf(r)}$ because a B&uuml;chi automaton accepts if, and only if, some accepting state is visited infintely often. Let $s_0$&nbsp;be any prefix of&nbsp;$s$ such that $A$&nbsp;is in state&nbsp;$q$ after reading&nbsp;$s_0$.  Since $r$ visits&nbsp;$q$ infinitely often, it must return to&nbsp;$q$ when processing&nbsp;$s$, say after processing the prefix&nbsp;$s_0s_1$.  In particular, if the automaton is in state&nbsp;$q$ and reads string&nbsp;$s_1$, there is a run in which it returns to&nbsp;$q$. Therefore, the automaton has a run that visits&nbsp;$q$ infinitely often when processing the string $s_0\,{s_1}^\omega = s_0s_1s_1s_1\dots$.