Is there a regular not completely regular space which is not a corkscrew? While studying some examples of regular spaces which are not completely regular, I came across Steen's and Seebach's "Counterexamples in Topology". In this book, after searching for examples, I only found four examples of regular not completely regular spaces: Tychonoff Corkscrew, deleted Tychonoff Corkscrew, Hewitt's Condensed Corkscrew and Thomas' Corkscrew.
All this examples are variations of the same example: the Tychonoff Corkscrew. So my question is the following one: is there a regular not completely regular space which is not a corkscrew? Or in other way, is it possible to define the property of being a corkscrew is some sense in order to have a result of the form that every regular space is completely regular except if it is a corkscrew?
Alternatively, is it possible to construct a regular not completely regular space which is not a corkscrew in the sense of being radically different -in the sense that use totally different ideas and it is not homeomorphic to the cited ones- from the previous examples?
 A: The following example (which I am fairly certain is not a corkscrew-like construction) is essentially taken from

A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), pp.652-653, MR601748, AMS link

Let $X = (\;\mathbb{R} \times [ 0 , 2 )\;) \cup \{ \langle 0 , -1 \rangle \}$, and topologise $X$ as follows:


*

*all points in $\mathbb{R} \times ( 0 , 2 )$ are isolated;

*for $\langle x , 0 \rangle$ the basic open neighbourhoods are of the form $V_x \setminus A$ where $V_x = \{ \langle x,y \rangle : 0 \leq y \leq 2 \} \cup \{ \langle x + y , y \rangle : 0 \leq y < 2 \}$, and $A \subseteq V_x \setminus \{ \langle x,0 \rangle \}$ is finite; and

*the basic open neighbourhoods of $\langle 0 , -1 \rangle$ are of the form $U_n = \{ \langle 0 , -1 \rangle \} \cup \{ \langle x , y \rangle \in \mathbb{R} \times [0,2) : x > n \}$ for $n \in \mathbb{N}$.


It is fairly easy to check that $X$ is regular, the basic idea being that the basic open neighbourhoods of the points in $\mathbb{R} \times [ 0 , 2 )$ described above are actually clopen, and $\overline{U_{n+2}} \subseteq U_n$ for all $n \in \mathbb{N}$.
However there is no continuous $f : X \to [0,1]$ such that $f(0,-1) = 0$ and $F = \{ \langle x , 0 \rangle \in X : 0 \leq x \leq 1 \} \subseteq f^{-1} [ \{ 1 \} ]$.  The basic idea here is to show that if $F \subseteq f^{-1} [ \{ 1 \} ]$, then $f( 0 , -1 ) = 1$.  For this we first prove by induction that $B_n = \{ x \in ( n-1 , n ] : f( x , 0  ) = 1 \}$ is infinite for all $n \geq 1$.

For $n = 1$ the result is trivial.  So suppose that $B_n$ is infinite for some $n \geq 1$.  For each $x \in B_n$, one can show that there is a countable $C_x \subseteq [0,2)$ such that $f( x + y , y ) = 1$ for all $y \in [ 0 , 2 ) \setminus C_x$.  Fixing a countably infinite $B^\prime \subseteq B_n$, it follows that $C = \bigcup_{x \in B^\prime} \{ x + y : y \in C_x \}$ is countable, and so $(n,n+1] \setminus C$ is (uncountably) infinite.  For $z \in [n,n+1] \setminus C$, note that for each $x \in B^\prime$ we have that $z = x + ( z - x ) \notin C_x$, and so $f( z , z-x ) = 1$.  Since there are infinitely many $\langle z,y \rangle \in V_z$ such that $f(z,y) = 0$, by continuity it follows that $f ( z, 0 ) = 1$.  Therefore $B_{n+1} \supseteq (n,n+1] \setminus C$ is infinite.

Now note that if $f ( 0,-1 ) \neq 1$, then there would be an $n \geq 1$ such that $U_n \subseteq f^{-1} [\;[0,1)\;]$, however as $B_{n+1} \subseteq U_n$ this is impossible.
