[proofplan] The $T_1$ hypothesis turns point separation into closed-set separation: for distinct points $x,y \in X$, the singleton sets $\{x\}$ and $\{y\}$ are closed. Since the points are distinct, these two closed sets are disjoint. Normality then supplies disjoint open neighborhoods of the two singleton sets, which are precisely the Hausdorff neighborhoods required. [/proofplan] [step:Convert the two distinct points into disjoint closed singleton sets] Let $x,y \in X$ be distinct points. Define \begin{align*} A := \{x\}, \qquad B := \{y\}. \end{align*} Since $(X,\tau)$ is $T_1$, every singleton subset of $X$ is closed. Hence $A$ and $B$ are closed subsets of $X$. Since $x \neq y$, no point of $X$ belongs to both $A$ and $B$, so \begin{align*} A \cap B = \varnothing. \end{align*} [guided] We begin with arbitrary distinct points $x,y \in X$ because the Hausdorff condition must be proved for every such pair. The goal is to use normality, but normality separates closed sets, not points directly. Therefore we replace the points by the singleton sets \begin{align*} A := \{x\}, \qquad B := \{y\}. \end{align*} The $T_1$ hypothesis is exactly what permits this replacement: in a $T_1$ [topological space](/page/Topological%20Space), every singleton subset is closed. Applying this to the points $x$ and $y$, we obtain that $A$ and $B$ are closed subsets of $X$. It remains to check that these two closed sets are disjoint, since normality applies only to disjoint closed subsets. If $z \in A \cap B$, then $z=x$ because $z \in A=\{x\}$, and $z=y$ because $z \in B=\{y\}$. Thus $x=y$, contradicting the assumption that $x$ and $y$ are distinct. Therefore \begin{align*} A \cap B = \varnothing. \end{align*} [/guided] [/step] [step:Apply normality to separate the singleton sets by disjoint open neighborhoods] By normality of $(X,\tau)$, any two disjoint closed subsets of $X$ have disjoint open neighborhoods. Applying this to the disjoint closed subsets $A$ and $B$, there exist open sets $U,V \in \tau$ such that \begin{align*} A \subset U, \qquad B \subset V, \qquad U \cap V = \varnothing. \end{align*} Since $x \in A$ and $y \in B$, it follows that \begin{align*} x \in U, \qquad y \in V, \qquad U \cap V = \varnothing. \end{align*} Thus every pair of distinct points in $X$ has disjoint open neighborhoods, so $(X,\tau)$ is Hausdorff. [/step]