In real analysis, we learned that $x \in cl(A)$ iff $\exists$ a sequence $x_n \to x$ with $x_n \in A$.
I have to show that in general this statement is not true, so I was trying to use the co-countable where the open sets are those whose complement is countable. Is the following argument sound?
Let $X = \mathbb{R}, A = \mathbb{R} \setminus \{p\}$, where $X$ is equipped with the co-countable topology. Clearly $p$ is in the $cl(A)$, which means if the analytic definition of closure holds, we can find a sequence $x_n \to p$ with $x_n \in A$. In a topological space, we have $x_n \to p$ if $\forall U \in \tau_X$ with $p \in U$, we can find $N \in \mathbb{N}$ such that $x_n \in U$ for $n \geq N$. Consider $U = \mathbb{R} \setminus \{x_n | x_n \neq p\}$, since the sequence $x_n$ is countable, $U \in \tau_X$. Since by construction $p \in U$, there exists an $N \in \mathbb{N}$ for which $x_n \in U$ for $n \geq N$. However, all $x_n \neq p$ are not in $U$, so $x_n = p$ for $n \geq N$ and is eventually constant. But this is a contradiction, since $p \not \in A$, there exist $x_n \not \in A$.
