In the book of Topology by Munkres, at page 193, it is given that
Definition. Suppose that one-point sets are closed in X. Then X is said to be regular if for each pair consisting of a point x and a closed set B disjoint from x, there exist disjoint open sets containing x and B, respectively. The space X is said to be normal if for each pair A, B of disjoint closed sets of X, there exist disjoint open sets containing A and B, respectively.
It is clear that a regular space is Hausdorff, and that a normal space is regular. (We need to include the condition that one-point sets be closed as part of the definition of regularity and normality in order for this to be the case. A two-point space in the indiscrete topology satisfies the other part of the definitions of regularity and normality, even though it is not Hausdorff.)
But, how does assuming one-point sets are closed allow regular (normal) space to be Hausdorff(regular) ?
Further considerations:
(i). Consider Sierpinski space $S=\{x,y\}$ with $x\ne y,$ where $S,\emptyset,$ and $\{x\}$ are open but $\{y\}$ is not open. If $A, B$ are disjoint closed subsets of $S$ then at least one of $A, B$ is empty so $A,B$ are covered by disjoint open sets. But $S$ does not satisfy the condition for regularity: $x$ does not belong to the closed subset $\{y\}$ but the only open set covering $\{y\}$ is the whole space $S$.
(ii). Let $X$ be a normal space. Let $p\in X$ and let $Y$ be a closed subset of $X$ with $p\not \in Y.$ Since $X$ is a $T_1$ space (one-point subsets are closed), the sets $\{p\}, Y$ are closed and disjoint. So, since $X$ is normal, there are disjoint open sets with $\{p\}\subset U$ and $Y\subset V. $ That is, $p\in U$ and $Y\subset V.$ So $X$ is regular.