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

Let E be a co-countable space and E be uncountable.
To show: A sequence in E is convergent iff the sequence is eventually constant.
If the sequence is eventually constant, then it converges to that constant.
Let $S = \{x_{n}\}_{n\geq 1} $ be a convergent sequence converges to $x$ in E.
Suppose S is not eventually constant, then for all $n\in \mathbb{N} $ there exists $m > n$ such that $x_{m} \neq x_{n}$.
Let n= 1, there exists $m_{1} > 1$ such that $x_{1} \neq x_{m_{1}}$ and 
     for $m_{1}$ there exists $m_{2} > m_{1}$ such that $ x_{m_{1}} \neq x_{m_{2}} $ we can continue like this. So we get a subsequence $T =\{x_{m_{k}} \}_{k \geq 1} $ such that $ x_{m_{i}} \neq x_{m_{j}} $, if $i \neq j$.
Since T is a subsequence of S , hence converges to $x$. 
Suppose x is not in T, then x is in complement of T which is a open set contradicts that T is a convergent sequence converging to x.
So, now let $x_{m_{l}} = x$ for some l. Now $x_{m_{l+k}}$ for all $l \geq 1$ is a subsequence of T which converges to x. Again by previous argument we can say that $ x_{m_{l+l_{0}}} = x$, for some  $l_{0}$. This is a contradiction to the choice of T.
