Property of Hausdorff spaces I want to show that $$\text{E is a Hausdorff space} \iff \bigcap_{V(x) \text{ is a closed neighborhood of X}} V(x)=\{x\}$$
I am sure my proof is incorrect, since along the way, I have managed to proove that $$\text{E is a Hausdorff space} \iff \bigcap_{V(x) \text{is a neighborhood of X}} V(x)=\{x\}$$ and it must be a $\implies$ only...
Here is what I did :
$\begin{aligned}
\text{E is not a Hausdorff space} & \iff \forall V \in V(x) \quad \forall U\in V(Y), \quad U\cap V \neq \varnothing \\
& \iff \exists z\in E : z\in U\cap V \quad \forall V \in V(x) \quad \forall U\in V(Y) \\
& \iff \exists z\in E : z\in V \quad \forall V \in V(x) \\
& \iff \exists z\in E : z\in \bigcap V \quad \text{where V is a neighborhood of }x \\
& \iff \exists z\in E : z\in \bigcap V_F \quad \text{where $V_F$ is a closed neighborhood of }x \\
& \iff \text{the intersection of all neighborhoods of }x \\
& \phantom{\iff} \ \text{ 
does contain another element than }x 
\end{aligned}$
Which proves the statement.
Where did I make a mistake ?

Taking different $U$ and $V$ may give you a different $z$. By assuming that the $z$ is the same, you assume the wanted conclusion.
