Unique limits of sequences plus what implies Hausdorff?

First countable is enough. Let $x\neq y$ be two points in your space that cannot be separated by neighborhoods. Let $O_1,O_2,\ldots$ form a neighborhood base of $x$ and let $U_1,U_2,\ldots$ form a neighborhood base for $y$. Choose a sequence $(z_n)$ such that $z_n\in O_n\cap U_n$ for all $n$. Now $(z_n)$ converges to both $x$ and $y$.

Here is an answer to Dirk's last question, "Is there a class of non_Hausdorff spaces in which convergent sequences have unique limits?"

Yes. The so called KC-spaces or maximal compact spaces. These are spaces such that every compact subspace is closed.

(The 1967 Monthly article of Wilansky, "Between $\mathrm T_1$ and $\mathrm T_2$" (MSN), subsumes, references, or implies all of the following.)

In a KC-space, convergent sequences have unique limits.

(Suppose $x_n\to x$ in the KC space $X$. The set $\{x,x_1,x_2,\dotsc\}$ is compact and hence closed. Thus, if $y$ is not in the set $\{x,x_1,x_2,\dotsc\}$, then the open set $X \setminus \{x,x_1,x_2,\dotsc\}$ shows it is false that $x_n\to y$. Thus, if $x_n\to y$, then $y=x$ or $y=x_n$ for some $n$. If $y=x_n$ for infinitely many indices $n$ then $y=x$ (since every KC space is $\mathrm T_1$ (since singletons are compact) and since constant sequences have unique limits in a $\mathrm T_1$ space). If $y=x_n$ for finitely many indices $n$ then (deleting $y$ from the sequence $x_1,x_2,\dotsc$) we are left with a subsequence $z_n\to x$, the knowledge that $y$ is not equal to any $z_n$, and the knowledge that $y$ is in the set $\{x,z_1,z_2,\dotsc\}$, and we conclude that $y=x$).

To exhibit a large class of non-Hausdorff KC spaces, let $X$ be a non-locally-compact metric space (for example, the rationals) and let $Y=X \cup \{y\}$ denote the Alexandroff compactification of $X$ (i.e., $V$ is open in $Y$ if $V$ is open in $X$ or if $Y\setminus V$ is a compact subspace of $X$).

The space $Y$ is a KC space, but $Y$ is not Hausdorff.

Here's an example of a space which is not Hausdorff but which has unique limits...

Let $X = \mathbb{R}$ with the cocountable topology, i.e. a set is open iff its complement is countable. Clearly any two open sets intersect, because $\mathbb{R}$ is uncountable. So $X$ is non-Hausdorff. Now, suppose $(x_n)$ is a sequence which converges to $x$. Then $C =$ {$x_n\;|\;x_n\neq x$} is closed because it's countable. So $X-C$ is a neighborhood of $x$ and this means there is some $N$ such that for all $n>N$ $x_n\in X-C$, i.e. $x_n=x$ for large $n$. This means if $x_n\rightarrow y$ then $y=x$, proving limits are unique.