Index the copies of $P^*$ by the integers, so that you have $P^*\times\Bbb Z$, and let $P^*_k=P^*\times\{k\}$, the $k$-th level of the space, for each $k\in\Bbb Z$. Denote by $\langle\pm\alpha,\pm n,k\rangle$ the copy of $\langle\pm\alpha,\pm n\rangle$ in $P^*_k$.
So far you just have a stack of unrelated copies of $P^*$. The point $\langle\alpha,\omega,k\rangle$ on level $k$ is the limit from the ‘north’ of points $\langle\alpha,n,k\rangle$ and from the ‘south’ of points $\langle\alpha,-n,k\rangle$, all in level $k$. Now you angle the fourth quadrant of $P^*_k$ down so that the points $\langle\alpha,-n,k\rangle$ converge to $\langle\alpha,\omega,k-1\rangle$ instead of to $\langle\alpha,\omega,k\rangle$. You leave the first quadrant alone, however, so that the points $\langle\alpha,n,k\rangle$ still converge to $\langle\alpha,\omega,k\rangle$. The picture is correct: if you start in the first quadrant of the upper level in the picture, you can walk west into the second quadrant, then south into the third, then east into the fourth, and if you now walk north, when you cross that lower dashed line you’ll find yourself in the first quadrant of the lower level.
I suspect that Steen and Seebach chose this version of the construction because it looks the most like a real corkscrew, but it does make it a bit more difficult to give a rigorous description. Here’s one way to do it. Let $P$ now be the deleted Tikhonov plank. Let $Y=P\times\Bbb Z$, and let $P_k=P\times\{k\}$ for $k\in\Bbb Z$. The sets $P_{4k}$ for $k\in\Bbb Z$ will be the first quadrants of the S&S version; the sets $P_{4k+1}$ will be the second quadrants, the sets $P_{4k+2}$ the third quadrants, and the sets $P_{4k+3}$ the fourth quadrants. Define an equivalence relation $\sim$ on $Y$ as follows: $y\sim y$ for each $y\in Y$, of course, and in addition
- $\langle\Omega,n,4k\rangle\sim\langle\Omega,n,4k+1\rangle$ for $k\in\Bbb Z$ and $0\le n<\omega$;
- $\langle\alpha,\omega,4k+1\rangle\sim\langle\alpha,\omega,4k+2\rangle$ for $k\in\Bbb Z$ and $0\le\alpha<\Omega$;
- $\langle\Omega,n,4k+2\rangle\sim\langle\Omega,n,4k+3\rangle$ for $k\in\Bbb Z$ and $0\le n<\omega$; and
- $\langle\alpha,\omega,4k+3\rangle\sim\langle\alpha,\omega,4k-4$ for $k\in\Bbb Z$ and $0\le\alpha<\Omega$.
Now let $X=Y/\sim$, the quotient space.
The first line of the definition of the non-trivial part of $\sim$ sews $P_{4k}$ to $P_{4k+1}$ along the line that looks like the positive $y$-axis in the S&S pictures. The second sews the ‘second quadrant’ $P_{4k+1}$ to the ‘third quadrant’ $P_{4k+2}$. The third sews the ‘third quadrant’ $P_{4k+2}$ to the ‘fourth quadrant’ $P_{4k+3}$. And the last sews the ‘fourth quadrant’ $P_{4k+3}$ not to the ‘first quadrant’ $P_{4k}$, where we started, but to $P_{4k-4}$, the ‘first quadrant’ one place lower in the stack. Thus, the corkscrew (without the ideal points called $a^+$ and $a^-$ in S&S) is just a quotient space of the product space $P\times\Bbb Z$.
A simpler example of a $T_3$-space that’s not Tikhonov was constructed by John Thomas; apart from two ideal points, it lives in $\Bbb R^2$, and I recommend making a sketch.
For each even $n\in\Bbb Z$ let $L_n=\{n\}\times\left[0,\frac12\right)$. For each odd integer $n$ and each integer $k\ge 2$ let $p_{n,k}=\left\langle n,1-\frac1k\right\rangle$, and let $$T_{n,k}=\left\{\left\langle n\pm t,1-t-\frac1k\right\rangle:0<t\le1-\frac1k\right\}\;.$$ $T_{n,k}$ consists of the legs of an isosceles right triangle with its apex at $p_{n,k}$; the base of the triangle is not part of the space, but it’s the segment of the $x$-axis from $\left\langle n-1+\frac1k,0\right\rangle$ to $\left\langle n+1-\frac1k,0\right\rangle$, inclusive. Let $$Y_0=\bigcup_{n\in\Bbb Z}L_{2n}\;,$$ $$Y_1=\{p_{2n+1,k}:n\in\Bbb Z\text{ and }k\ge 2\}\;,$$ and $$Y_2=\bigcup\{T_{2n+1,k}:n\in\Bbb Z\text{ and }k\ge 2\}\;.$$ Let $Y=Y_0\cup Y_1\cup Y_2$.
Points of $Y_2$ are isolated. A nbhd of $p_{n,k}$ must contain all but at most finitely many points of the ‘tent’ $T_{n,k}$ whose peak it is. And a nbhd of a point $\langle n,y\rangle\in L_n$ must contain all but at most finitely many points of the line segment $(n-1,n+1)\times\{y\}$. (The points on this segment are the points at height $y$ on the righthand sides of the tents $T_{n-1,k}$ and the lefthand sides of the tents $T_{n+1,k}$.
Now let $X=Y\cup\{p^-,p^+\}$, where $p^-$ and $p^+$ are distinct new points. For each $\alpha\in\Bbb R$, $\{p^-\}\cup\{\langle x,y\rangle\in Y:x<\alpha\}$ is a nbhd of $p^-$, and $\{p^+\}\cup\{\langle x,y\rangle\in Y:x>\alpha\}$ is a nbhd of $p^+$. You’ll need to check, but the nbhd assigments really do specify a topology on $X$. (This is essentially Thomas’s original presentation of the space in A Regular Space, Not Completely Regular, The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 181-182. S&S give a different presentation of it as their example $94$.
It’s not too hard to show that if $f:X\to\Bbb R$ is continuous, then $f(p^-)=f(p^+)$. The key step is observing that if $f(p_{n,k})=\alpha$, then there are at most countably many points $z\in T_{n,k}$ such that $f(z)\ne\alpha$.
