There is a theorem which says that every order topology is Hausdorff. Also every Hausdorff follows $T_1$ Axiom.
So suppose $X = \{1,2\}$. Now $1 < 2$. A basis for the topological space $X$ is $B_1=[a,b)$ and $B_2=(a,b]$. We have an open neighborhoods $B_1$ and $B_2$ around $1$ and $2$ respectively, such that $B_1 \cap B_2 = \emptyset$. So $X$ is a Hausdorff space.
But since it also satisfies the $T_1$ Axiom it means $B_1$ and $B_2$, which are singleton sets, are closed. That means $B_1$ and $B_2$ are not open neighborhoods.
If anyone can please tell where am I getting confused?