Sequence of points of $X$ converge to at most one point of $X$ I know that the Hausdorff axiom implies convergence to at most one point. Is the reverse true? If not, is there any relation between the separability axioms and convergence to at most one point?
 A: Consider the Co-Countable topology on $\Bbb R$, that is $U\subseteq \Bbb R$ is open iff either $U=\emptyset$ or $\Bbb R\backslash U$ is at most countable. This space is not Hausdorff. But still limit of each sequence is unique. 
To show this let $\{x_n\}$ be a sequence converging to two distinct points $x,y$. Then choose a nbd $V$ of $x$ that does not contain $y$, then $\{x_n:n\geq n_0\}\subseteq V$ for some $n_0\in \Bbb N$. So $W=\Bbb R\backslash\{x_n:n\geq n_0\}$ is nbd of $y$ not containing any tail of the sequence, contradiction.
If we assume our space $X$ with some topology is First-Countable as well as each sequence has unique limit point, then $X$ will be hausdorff. 
We can again prove this last assertion by contradiction. So let $X$ is first contable as well as each sequence has unique limit point, but $X$ is not hausdorff. So there two distinct points $x,y\in X$, such that every nbd of $x$ intersects with every nbd of $y$. But we assume $X$ is first countable. So there is a countable local base $\{U_n\}$ at $x$ and a countable local base $\{V_n\}$ at $y$. But, $U_n\cap V_n\not=\emptyset,\forall n\in \Bbb N$. Choose a point $z_n\in U_n\cap V_n$, then $\{z_n\}$ converges to both $x$ and $y$, contradicting to our assumption.
A: A space $X$ is called a US (unique sequential limit) space (from this paper) if whenever $(x_n)$ is a sequence in $X$ and $x_n \to x$ and $x_n \to y$ for $x,y \in X$, we can conclude that $x=y$. I.e. if a sequence converges in $X$, its limit is unique.
This implies that $X$ is $T_1$: for suppose that $X$ is not $T_1$ then we have $x \neq y$ such that $y$ is in the closure of $\{x\}$ (a set is $T_1$ iff all singleton sets are closed). This means that every open set that contains $y$ also contains $x$, and so the (constant) sequence defined by $x_n = x$ converges to $y$ and it also converges to $x$, while $x \neq y$, contradicting that $X$ is a US space. 
As you say, it's well known that $T_2$ (Hausdorff) spaces are all US spaces too. In fact, if we extend convergence to that of nets or filters, unicity of limits becomes equivalent to Hausdorffness, and that we can even have this weaker notion shows the inherent limitation of sequences, in a way. The cocountable topology on an uncountable set has the property that all non-empty open sets intersect (so it's very non-Hausdorff), but the only convergent sequences are the one that are eventually equal to their (unique!) limit. It shows that US is properly weaker than Hausdorff.
But the notion is not that important, and is not covered in any important or well-used text books. Even Engelking doesn't mention it, nor the Handbook of Set-theoretic topology (my often consulted references). But the question you ask is a natural one, and has been somewhat studied.
