Space in which All Convergent Sequences have Unique Limit not necessarily Hausdorff

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $T$ be such that all convergent sequences have a unique limit point.

Then it is not necessarily the case that $T$ is a Hausdorff space.

Proof
Let $T = \left({\R, \tau}\right)$ be the set of real numbers $\R$ with the countable complement topology.

From Countable Complement Space is not $T_2$, $T$ is not a Hausdorff space.

Suppose $\left\langle{x_n}\right\rangle$ is a sequence in $\R$ which converges to $x$.

Then $C = \left\{{x_n: x_n \ne x}\right\}$ is closed in $T$ because it is countable.

So $X \setminus C$ is a neighborhood of $x$.

This means there is some $N \in \N$ such that:
 * $\forall n > N: x_n \in X \setminus C$

That is, $x_n = x$ for large $n$.

This means that if $x_n \to y$ then $y = x$, proving limit points in $T$ are unique.