in $\mathbb{R}^2$ for any y take $B_y:=\{(x,y) \in\mathbb{R}^2 : x \in \mathbb{R}\}$. Consider the basis of the topology to be $$\mathcal{B}=\{B_y: y \in \mathbb{R}\}.$$
This gives a non-hausdorff topology since for any two points in the same vertical line any open set that contains one contains the other. Is this correct?
The sets of the basis are the lines parallel to $x-$axis.
Take any point $z_1=(x_1,y),z_2=(x_2,y)$ where $x_1 \neq x_2$ and $y \in \Bbb{R}$
There do not exist disjoint open sets $A,B$ such that $z_1 \in A$ and $z_2 \in B$
Note that from the definition of the basis, every open set in this topology is expressed as a(disjoint) union of the sets $B_y$