Sequence in Indiscrete Space converges to Every Point

Theorem
Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Let $\left \langle {s_k} \right \rangle$ be a sequence in $T$.

Then $\left \langle {s_k} \right \rangle$ converges to every point of $S$.

Hence if $S$ is uncountable, every sequence in $T$ has an uncountable number of limit points.

Proof
Let $\alpha \in S$.

By definition, $\left \langle {s_k} \right \rangle$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\left \langle {s_n} \right \rangle$.

But as $T$ has only one open set containing any points at all, every point of $\left \langle {s_n} \right \rangle$ is contained in every open set in $T$ containing $\alpha$.

Hence the result.