In the book "General topology" by Engelking, where is exercise to build regular topological space, for which there exist two distinct points, $a$ and $b$, for which is correct sentence: for every real valued continuous function $f(a)=f(b)$ (exercise 2.7.17. page 119).
There is also hint, that such space may build from adjoin the point for the space, which is build by A. MYSIOR, which is regular and not completely regular space. But, I still can`t prove that exercise. please. help me if you can.
Take Y as in the remark of the article by A.Mysior you have linked. I will follow its notations.
As it is said in the article, Y is regular.
Now take any real valued continuous function $f$ on Y.
For $\epsilon >0$, $\exists N \in \mathbb{N} $ such that $U_N(b) \subset f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$ (because since f is continuous, $f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$ is a neighborhood of $b$) .
Set $J_n = f^{-1}([f(b)- \epsilon , f(b) + \epsilon]) \cap \{(x,0) \ | \ n-1\leq x \leq n \}$, since $J_{-N-1}$ is infinite, you can apply the same inductive reasoning as in the article (where it is done for $K_n$), hence $J_n$ is infinite $\forall n \in \mathbb{Z}$.
By continuity, we get $a \in f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$.
Thus $f(a)=f(b)$.
I hope this can help.