Let $E$ be an uncountable set with $\tau:=$ co-countable topology.
WTP: A sequence in $E$ is convergent iff the sequence is eventually constant.
$"\Rightarrow"$ Assume $(x_n)_{n \in \Bbb N} \in E$ is a convergent sequence. So that is, $(x_n)_{n \in \Bbb N} \to l, l \in E$ Then, $\forall V \in \mathscr{N}_{\tau}(l), \exists N \in \Bbb N, n \ge N, x_n \in V$. Since $\tau$ is the co-countable topology $V^c$ is countable (or $V = \emptyset$). I can't figure out how to finish this one, more or less not sure how to put it down exactly. Seems the idea is that if the sequence is not uncountable at some point then it can't converge to something in the topological space.
$"\Leftarrow"$ Assume $(x_n)_{n \in \Bbb N}$ is eventually constant, then $\forall n \ge N$ we have $x_n \in \{x_N, x_{N+1}, \dots\} \wedge x_n = x_{N+i}, 0 \le i < \infty$ as $n \to \infty$. so $\forall V \in \mathscr{N}_{\tau}(x_N), \exists N \in \Bbb N,n \ge N, x_n \in V$ so it converges.
Best Answer
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.