Are there an completely regular, non-Lindelöf spaces with only constant real valued functions?

Question

Is there an example for a topological space, which is not Lindelöf, but is completely regular, on which continuous real valued functions, are, constant for all $x \in X$, or, constant for all, except finite sets in $X$?

Answer 1

If $X$ is completely regular, this means that for every open set $U$ and every $x \in U$ there exists some $f: X \rightarrow \mathbb{R}$ such that $f(x) = 1$ and $f[X \setminus U] = \{0\}$. Such a function is non-constant if $X \neq U$. So if $X$ is also $T_0$, for every $x \neq y$ we find some open set $U$ that contains one of them but not the other, and an $f$ as before will give a function such that $f[\{x,y\}] = \{0,1\}$; we can separate all points with different function values. Note that then also $f^{-1}[(-\infty,\frac{1}{2})]$ and $f^{-1}[(\frac{1}{2},+\infty)]$ are disjoint open sets around $x$ and $y$ so $X$ is then automatically Hausdorff as well.

So the only way I see for an infinite $X$ to be completely regular and all real-valued functions constant (except for possibly a finite set) is that $X$ is not even $T_0$ and has very few open sets. Why do you want such an example?