Büchi automata: accepting run vs. runs with arbitrarily many final states
Consider $\Sigma = \{0,1\}$ and $w = 1101001000100001\ldots$, i.e., there is an increasing number of zeros between two ones. Now consider the automaton with states $q_0, q_1$, and $q_A$ where $q_A$ is the only accepting state. On input $0$, always stay in the current state. On input $1$ allow the following transitions: In state $q_0$, either stay in $q_0$ or go to state $q_A$; in state $q_A$, go to $q_1$; in state $q_1$, stay in $q_1$.

If the transition $q_0 \to q_A$ happens at a position in $w$ that is followed by $n$ zeros, $q_A$ will occur exactly $n+1$ times in that run. Since $w$ contains $n$ consecutive zeros for all $n$ but no infinite sequence of zeros, this is a counterexample to your statement.