A sequence in a Hausdorff space and in a space that is not Hausdorff. 
Let $X$ be a topological space and $\{x_n\}_{n=1}^{\infty}$ a sequence in $X$.
Show that if $X$ is Hausdorff, $x_n \rightarrow x \:$, $x_n \rightarrow y \:$  implies $x=y$.
Give an example that shows that if $X$ is not Hausdorff then this is not necessarily true.

A sequence is convergent, $x_n \rightarrow x \:$, if there is an element $x \in X$ such that for every open neighborhood $U$ of $x$ there exists a $n_0$ such that $n > n_0$ implies $x_n \in U$.
If $X$ is Hausdorff and $x$ and $y$ are different there exists open disjoint sets $x \in U$ and $y \in V$.
Because all but a finite number of $x_n$ lies in $U$ according to the definition of that $x_n \rightarrow x \:$ then only a finite number of $x_n$ can be in $V$, which is a contradiction to that $x_n \rightarrow y \:$.
But i have a hard time to come up with an example that shows that this not necessarily is true when $X$ isnt Hausdorff.
Can anyone give me a hint?
Thanks.

Take $\mathbb{N}$ in the cofinite topology (the only closed sets are the finite ones (including the empty set) and $\mathbb{N}$ itself). Take $a_n$ to be any sequence where all values are different, like $a_n = n$ or $a_n = 2n$ etc.. Then $(a_n)$ converges to every point $m$ of $\mathbb{N}$, because the only open sets that contain $m$ are of the form $O = \mathbb{N} \setminus F$, where $F \subset \mathbb{N}$ is finite. But after some initial segment the values of $a_n$ are never in $F$ (as $F$ is only finite), so all values from some value onwards are in $O$. As this holds for all $O$, $(a_n) \rightarrow m$ for all $m$.
So any two of these sequences have all points of the space as their limit as well and so the limits are very non-unique. The cofinite topology is $T_1$, so one separation axiom below Hausdorff, making it a "sharper" example than the indiscrete topology. 
