Does T3 spaces implies every open set contains a closure of an open set I have two questions regarding this spaces:

First Question:  Suppose $(X,\tau)$ is T3. Does it implies that for every open set $U \subset X$ and for every $x \in U$ there exist an open set $V$, s.t. $x \in V \subseteq U$ and $V \subseteq \overline{V} \subseteq U$?
It seems to me that this is true since: 
Take $U$ open and $x \in U$. Then $U^c$ is closed. $X$ is T3 so there are two open disjoint sets $V_1,V_2$ s.t. $x \in V_1$ , $U^c \subset V_2$. So $V_1$ is the set we are looking for. since  $\overline{V_1} \subseteq {V_2}^c \subseteq U$ 

Am I right?

Second question: I am looking for an example for a space which is T3 but not T0,T1,T2.
I have looked in "Counter examples in topology" examp. 90, but I don't know what Tychonoff Corkskrew is..

Thank you!! Shir
The answer to your first question is yes, and your proof is correct. (This is sometimes even taken as the definition of regularity.)

As for your second question, if you use what I consider the standard definition of $T_3$, every $T_3$ space is automatically $T_0,T_1$, and $T_2$. To me a $T_3$ space is one that is regular (points can be separated from closed sets not containing them) and $T_1$, and such spaces are automatically $T_2$. Some people use $T_3$ to mean what I call regular, which is just the property that points can be separated from closed sets not containing them. In that case you can let $X=\Bbb R\cup\{p\}$, where $p$ is some point not in $\Bbb R$, with the following topology $\tau$:

$$\tau=\{U\subseteq\Bbb R:0\notin U\in\mathscr{E}\}\cup\{U\cup\{p\}:0\in U\in\mathscr{E}\}\;,$$

where $\mathscr{E}$ is the usual Euclidean topology on $\Bbb R$. This simply makes $p$ a second copy of $0$, so that $X$ is obtained from $\Bbb R$ by splitting $0$ into $0$ and $p$ and keeping both of these in the same open sets. Clearly $X$ is not $T_0,T_1$, or $T_2$, since neither of the points $p$ and $0$ can be separated from the other, but the space is still regular.

Your mention of the Tikhonov corkscrew suggests that you may instead be looking for an example of a $T_3$ space (in my sense) that is not Tikhonov. If so, you may find this answer helpful: it explains in more detail part of the construction of the corkscrew given by Steen &amp; Seebach and presents fully a simpler example of a $T_3$ space that isn’t Tikhonov.
A simple answer to your question is any Set X with at least two elements with the indiscrete topology T={empty set, X}, then it is clear that (X, T) is T3 (with respect to defintion of T3 in the book "Counter examples in topology") but it is neither To nor T1 nor T2
best regards 
Dr. Halgwrd M. Darwesh
Assistant Professor 
in General Topology and Dimensions Theory 
Math. Dep. , College of Sci. University of Sulaimani