Convergence in topology finer than cocountable

If $(X,\mathcal{T})$ is finer than cocountable topology then convergence is trivial (convergent sequences are only the eventually constant).

I already proved that convergence in cocountable topology is trivial, but how can I extend this result to finer topologies?

Suppose that $\lim_{n\to\infty}x_n=x$ and that $(x_n)_{n\in\Bbb N}$ is not eventually constant. Then the set $\{x_n\mid n\in\Bbb N\wedge x_n\ne x\}$ is a countable set. Let $O=\{x_n\mid n\in\Bbb N\wedge x_n\ne x\}^\complement$. Then $O$ belongs to the cocountable topology, and therefore $O\in\mathcal T$. Since $x\in O$, $O$ is then a neighborhood of $x$. But there are infinitely many $n$'s that don't belong to $O$ (since they belong to $\{x_n\mid n\in\Bbb N\wedge x_n\ne x\}$). This is impossible, since we are assuming that $\lim_{n\to\infty}x_n=x$.