The following four properties are equivalent:
(1). $Y$ is Hausdorff
(2). Let $p\in Y$. For each $q\neq p$, there is a nbhd $U(p)$ such that $q\notin \overline{U(p)}$.
(3). For each $p\in Y$, $\bigcap \{\overline{U}|U$ is a nbhd of $p\}=\{p\}$
(4). The diagonal $\Delta =\{(y,y)| y\in Y\}$ is closed in $Y\times Y$.
Now there are lots of way to prove this theorem, like $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4) \Rightarrow (1)$, I think this is the least number of implication needed to prove this theorem(Note I’m not saying it is the most easiest way to do it) and $(1)\Leftrightarrow (2)$, $(1) \Leftrightarrow (3)$, $(1)\Leftrightarrow (4)$, this is maybe the most number of implication needed to prove this theorem. I want to prove this theorem by $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$ and $(1)\Leftrightarrow (4)$, because here is proof of $(1)\Leftrightarrow (4)$.
$(1)\Rightarrow (2)$. Let $p,q \in Y$ such that $p\neq q$. since $Y$ is Hausdorff, $\exists U,V\in \mathcal{T}_Y$ such that $p\in U$, $q\in V$ and $U\cap V=\emptyset$. Which implies $U\subseteq Y-V$. By Exercise 6, Section 17 of Munkres’ Topology, $\overline{U}\subseteq \overline{Y-V}=Y-V$. So $V\subseteq Y-\overline{U}$. $q\in V \Rightarrow q\in Y-\overline{U}$. $q\notin \overline{U}$. Thus $\exists U\in \mathcal{N}_p$ such that $q\notin \overline{U}$.
$(2)\Rightarrow (3)$. Let $p\in Y$. $\{p\}\subseteq U\subseteq \overline{U}$, $\forall U\in \mathcal{N}_p$. So $\{p\} \subseteq \bigcap \{ \overline{U}| U \in \mathcal{N}_p\}$. Conversely, assume towards contradiction, $\exists q\in Y-\{p\}$ such that $q\in \bigcap \{ \overline{U}| U \in \mathcal{N}_p\}$. So $p\neq q$. By $(2)$, $\exists W\in \mathcal{N}_p$ such that $q\notin \overline{W}$. So $q\notin \bigcap \{ \overline{U}| U \in \mathcal{N}_p\}$. Which contradicts our initial assumption. Thus $(\bigcap \{ \overline{U}| U \in \mathcal{N}_p\}) \cap (Y-\{p\})=\emptyset$. Which implies $\bigcap \{ \overline{U}| U \in \mathcal{N}_p\} \subseteq \{p\}$. Hence $\bigcap \{ \overline{U}| U \in \mathcal{N}_p\}=\{p\}$.
$(3)\Rightarrow (1)$. Let $p,q\in Y$ such that $p\neq q$. By $(3)$, $\{p\}= \bigcap \{ \overline{U}| U \in \mathcal{N}_p\}$. So $q\notin \bigcap \{ \overline{U}| U \in \mathcal{N}_p\}$. Which implies $\exists W\in \mathcal{N}_p$ such that $q\notin \overline{W}$. $Y-\overline{W} \in \mathcal{T}_Y$ and $q\in Y-\overline{W}$. Since $W\subseteq \overline{W}$, we have $(Y-\overline{W}) \cap W =\emptyset$. Thus $\exists W, Y-\overline{W} \in \mathcal{T}_Y$ such that $p\in W$, $q\in Y-\overline{W}$ and $(Y-\overline{W})\cap W=\emptyset$. Hence $Y$ is Hausdorff.
Is this proof correct?
