We say that a topological space $(X,\mathcal{T})$ is normal if any two disjoint closed subsets of $X$ are separated by neighbourhoods.
We have that in $(\mathbb{R},\{\emptyset,\mathbb{R}\})$ are $$\emptyset,\mathbb{R}$$ closed subsets, and themselves are neighbourhoods with obviously $$\emptyset \cap \mathbb{R} = \emptyset$$
So $(\mathbb{R},\{\emptyset,\mathbb{R}\})$ is a normal space, but $(\mathbb{R},\{\emptyset,\mathbb{R}\})$ is not a Hausdorff space.
This is a contradiction? Doesn`t normal ($T_4$) implies Hausdorff ($T_2$)?
Regards!
Best Answer
Normal doesn't actually imply Hausdorff; a space which is normal and $T_1$ is Hausdorff. Your example is a counterexample to the claim that every normal space is Hausdorff; note that your space is not $T_1$, since no single point is closed.
(It was pointed out in a comment that there's some inconsistency as to what normal and $T_4$ mean; I'm using the definition of normal that you gave, which doesn't include $T_1$.)