Hausdorff property in terms of closed sets Let $(X,\tau)$ be a topological space. Let us recall the two following definitions:

$(1)$ We say that $(X,\tau)$ is a $T_1$ (Fréchet) space iff for every $x_1,x_2 \in X$ there exist two open sets $A_1, A_2\in \tau$ such that $x_1\in A_1$, $x_2 \not \in A_1$, $x_1 \not \in A_2$, $x_2 \in A_2$.
$(2)$ We say that $(X,\tau)$ is a $T_2$ (Hausdorff) space iff for every $x_1, x_2\in X$ there exist two open sets $A_1,A_2\in \tau$ such that $A_1 \cap A_2=\varnothing$, $x_1\in A_1$, $x_2 \in A_2$.

In my lecture notes, the previous properties are also characterized through closed sets. The author states

$(1.C)$ $(X,\tau)$ is $T_1$ iff $\{x\}$ is closed for every $x\in X$ or, equivalently, given two points $x_1,x_2\in X$ we have that $x_2\not \in \overline{\{x_1\}}$ and $x_1 \not \in \overline{\{x_2\}}$.
$(2.C)$ $(X,\tau)$ is $T_2$ iff for every $x \in X$ the intersection of closed neighbourhoods of $x$ is $\{x\}$.

Now, the closure $\overline S$ of a set $S$ is defined to be the intersection of all closed sets containing $S$.
Question : if we restrict this definition to $S=\{x\}$, isn't the closure $\overline{\{x\}}$ the intersection of closed neighbourhoods of $x$? Isn't this property already implied by the fact that $\{x\}$ is closed, so $\overline{\{x\}}=\{x\}$  as stated in $1.C$?
 A: Let $X$ be an infinite set, and let $\tau$ be the cofinite topology on $X$. This topology is $T_1$ but not $T_2$. The closed sets in $X$ are the finite sets and $X$ itself, and of these, only $X$ has non-empty interior. Thus, for each $x\in X$ the only closed nbhd of $x$ is $X$ itself, and the intersection of the closed nbhds of $x$ is $X$, not $\{x\}$. (Recall that a set $C$ is a closed nbhd of $x$ if and only if $C$ is closed, and $x\in\operatorname{int}C$.) 
On the other hand, if $y\in X\setminus\{x\}$, then $X\setminus\{y\}$ is an open nbhd of $x$ that does not contain $y$, so the intersection of all open nbhds of $x$ is indeed $\{x\}$.
When $X$ is $T_2$, we can improve the argument in the last paragraph. If $y\in X\setminus\{x\}$, there are disjoint open sets $U_y$ and $V_y$ such that $y\in U_y$ and $x\in V_y$. Let $C_y=X\setminus U_y$; then $C_y$ is closed, and $x\in V_y\subseteq C_y$, so $x\in\operatorname{int}C_y$, and $C_y$ is therefore a closed nbhd of $x$ that does not contain $y$. Clearly, then,
$$\{x\}=\bigcap_{y\in X\setminus\{x\}}C_y\;,$$
so the intersection of all of the closed nbhds of $x$ must also be $\{x\}$.
