Convergent sequence in co-countable topology iff sequence is eventually constant

Let $(x_n)_{n\in \mathbb{N}} \subseteq E$ be a convergent sequence and let $l\in E$ such that $x_n \rightarrow l$. Define $V:=E\setminus \{ x_n : x_n \neq l\}$. Clearly $V$ has countable complement and is therefore open in our topology. As $x_n \rightarrow l$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ holds $x_n \in V$. By definition of $V$ this means $x_n = l$ for all $n\geq N$, i.e. $(x_n)_{n\in \mathbb{N}}$ is eventually constant.

Let $(x_n)_{n\in \mathbb{N}} \subseteq E$ be eventually constant. Then there exists $l\in E$ and $N \in \mathbb{N}$ such that for all $n\geq N$ holds $x_n=l$. Let $V\subseteq E$ be an open neighborhood of $l$. By definition of neighborhood $l\in V$. Hence, for all $n\geq N$ holds $x_n\in V$. As $V$ was an arbitrary neighborhood of $l$ we conclude $x_n \rightarrow l$. Note that we didn't use any information about the topology, i.e. this holds in every topological space.