Bases of completely regular (Tychonoff) spaces If the space $X$ is completely regular, we know that the collection 
{${\rm int}\,Z(f)$:$f$ is a continuous function from $X$ to the real numbers} is an open base for open subsets of the space $X$ (i.e., if for each element $x$  and each open set $U_x$ of $X$, there exist a continuous real-valued function $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\, Z(f)\subseteq Z(f)\subseteq U_x)$.
I have two questions about converse of this theorem. these questions are almost the same, but I think these are different.


*

*If for each element $x$  and each open set $U_x$ of $X$, there exist a continuous real valued function $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\, Z(f)\subseteq U_x$, then $X$ is completely regular.

*If for each element $x$  and each open set $U_x$ of $X$, there exist a continuous real valued function $f\colon X\to \mathbb{R}$ such that $x\in {\rm int}\,Z(f)\subseteq Z(f) \subseteq U_x$, then $X$ is completely regular.
I think these two claims have counterexamples and these conditions don't imply the complete regularity of $X$.
 A: Here's a counterexample to 1.
Let $T$ be the Tychonoff plank, i.e., the product 
$(\omega_1+1)\times(\omega+1)$ with the point $\langle\omega_1,\omega\rangle$
removed.
Consider the set $\omega\times T\cup\lbrace \infty\rbrace $ topologized so that
$\omega\times T$ has the product topology and is an open subset itself,
and the basic neighbourhoods of $\infty$ are of the form
$U_n(\infty) = (\omega\setminus n)\times T\cup\lbrace \infty\rbrace $.
We construct a quotient space by identifying
$\langle n,\alpha,\omega\rangle$ and $\langle n+1,\alpha,\omega\rangle$
whenever $n$ is odd and $\alpha\in\omega_1$, and
identifying
$\langle n,\omega_1,i\rangle$ and $\langle n+1,\omega_1,i\rangle$
whenever $n$ is even and $i\in\omega_1$.
The resulting space $C$, the Tychonoff corkscreww, is regular but not 
completely regular (the copy of $\lbrace 0\rbrace \times T$ and $\infty$ cannot
be separated by continuous functions).
For each odd $n$ define $f_n:C\to[0,1]$ by
$f_n(\infty)=0$ and
$$
f(m,\alpha,i)=
\begin{cases}
2^{-i} &\text{ if } m\le n \text{ and }i<\omega\cr  
0      &\text{ if } m> n \text{ and }i<\omega   
\end{cases}
$$
and $f_n(m,\alpha,\omega)=0$ for all $m$ and $\alpha$.
Then the interiors $\operatorname{int}Z(f_n)$ form a local base at $\infty$. 
