Definition: The topological space $X$ is $T_1$ if $ \{ x \} $ is closed for each point $x \in X$.
Proposition: The topological space $X$ is $\textit{Hausdorff}$ if and only if for each point $x \in X$ the intersection of the closed neighbourhoods of $x$ is the one-point set $\{ x \}$.
Question: Let $X$ be a $T_1$ space. Then for every $x\in X$, $ \{x \} $ is closed and $ \overline{\{ x \}} = \{ x\} $. Thus $ \{ x \} $ is the smallest closed set containing $ \{ x \} $, that is $ \{ x \} $ is equal to the intersection of all closed sets containing $ \{ x\} $. By the above proposition $ X $ is Hausdorff. Therefore every $T_1$ space is also Hausdorff.
But this statement is said to be wrong. Where have I made my mistake?
Thanks in advance!
The error lies in the assumption that, since $\{x\}$ is the intersection of all sets containing $x$, then it is also the intersection of all closed neighborhoods of $x$.
As an example, consider $\mathbb R$ with the topology $\tau$ for wich the open sets are $\emptyset$ and every $A\subset\mathbb R$ such that $A^\complement$ is finite. Then $(\mathbb{R},\tau$) is $T_1$, but not Hausdorff. In this case, the only closed neighborhood of a point $x$ is $\mathbb R$ itself.