Let $X = \mathbb{R}$ with the cocountable topology, i.e. a set is open iff its complement is uncountablecountable. 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.
Let $X = \mathbb{R}$ with the cocountable topology, i.e. a set is open iff its complement is uncountable. 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.