A $T_1$ space which is not Hausdorff More precisely, assume the following definitions.

Definitions. Let $S$ be a topological space. 
  
  
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*$S$ is a $T_1$ space if, whenever $s_1 \neq s_2$ there exists an open set $U_1$ such that $s_1 \in U_1$ but $s_2 \not\in U_1$, and there exists an open set $U_2$ such that $s_2 \in U_2$ but $s_1 \not\in U_2$.
  
*S is a $T_2$ space or Hausdorff space if, whenever $s_1 \neq s_2$, there exist open sets $U_j$ with $s_j \in U_j$ ($j = 1,2$ ) such that $U_1 \cap U_2 = \varnothing$; that is, $U_1$ and $U_2$ are disjoint.
  

I'm looking for a $T_1$ space which is not $T_2$. I know that metric spaces are Hausdorff (and even normal), so I discarded them. Moreover, topological spaces with at least two points and trivial topology are not Hausdorff but are not $T_1$ too. Topological spaces of the form $(S, \tau)$, with $S = \{ s_1, s_2 \}$, $\tau = \{ \varnothing, S, \{ s_1 \} \}$ are not $T_1$. (and hence are not Hausdorff.) Nevertheless, I'm sure it can't be a complicated stuff, since this exercise is assigned immediately below the above definitions.
Thanks for help!

Take the natural numbers with the cofinite topology, i.e., $U\subseteq \mathbb N$ is open iff $\mathbb N\setminus U$ is finite or $U=\varnothing$. 
