Every $T_3$ space is $T_2$ 
Prove that every $T_3$ space is $T_2$.

My attempt:
Suppose $x \neq y$. Then $x \notin \{y\} = \operatorname{cl}(\{y\})$, because $T_3$ means $T_1$, which is equivalent to saying that singeltons are closed.
Because a $T_3$ space is regular, we can find disjoint neighborhoods $U$ of $x$ and $V$ of {y}. But a neighborhood of $\{y\}$, is simply a neighborhood of $y$. Hence, we can find neighborhoods that separate the points $x,y$. Hence, the space is Hausdorff.
Is this correct?
 A: Usually, $T_3$ space are $T_2$ by definition: A space is $T_3$ if it is both regular and Hausdorff (see Wikipedia article on regular spaces)
If you are interested in an example of a regular, non Hausdorff space, here is one:
Add a point $*$ to $\mathbb R$: $X=\mathbb R\cup\{*\}$. Given $\tau$ the standard topology on $\mathbb R$, on $X$ put the topology $\sigma$:
$\sigma=\{A\in\tau: 0\notin A\}\cup\{A\cup\{*\}: A\in\tau, 0\in A\}$
In other words, the new point is topologicaly indistiguishable from $0$.
Note that an open/closed set contains $0$ if and only if it contains $*$.
$(X,\sigma)$ is clearly not $T_2$ as $0$ and $*$ have no disjoint neighborhoods. 
Let's see that $(X,\sigma)$ is regular. Let $x\in X$ and $A\subseteq X$ closed with $x\notin X$. (Note that $A\neq \{*\}$ because colsed sets containing $*$ contain also $0$.) 
If $x\neq *$ then $x$ and $A\setminus\{*\}$ are separated by open sets in $\tau$. If one of them contains $0$, then add $*$ to it. This results in two open sets of $\sigma$ separating $x$ and $A$.
If $x=*$ then $0\notin A$. Then, $A$ and $0$ are separated by open sets in $\tau$. Add $*$ to the open set containing $0$. This results in two open sets of $\sigma$ separating $x$ and $A$.
