= Solution
Set $S=\{x_n:n\in\mathbb N,\ x_n\ne\ell\}$. This is a countable <set>, hence a <closed set> in the stated <topology>, and $\ell\notin S$. Consequently $X\setminus S$ is an <open neighbourhood> of $\ell$. By <convergence of a sequence in a topological space>[convergence of a sequence], all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is \b[eventually constant], and only finitely many indices have $x_n\ne\ell$.
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