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.

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.