 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, all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is eventually constant, and only finitely many indices have $x_n\ne \ell$.