A nbhd base at a point $x$ in $X$ is a subcollection $\mathscr{B}_x$ taken from the collection of neighborhoods of $x$, $\mathscr{U}_x$, having the property that each $U \in \mathscr{U}_x$ contains some $V \in \mathscr{B}_x$.
A zero-set in a topological space $X$ is a set of the form $f^{-1}(0)$ for some continuous $f:X \to \mathbb{R}$
We want to show that a topological space $X$ is completely regular if and only if the zero set nbhds of each point form a nbhd base.
I'm not entirely sure where to start here for either direction - to be honest. I know that every closed subset of $\mathbb{R}$ is a zero set. But I'm not sure what to do beyond this point. But I'll try anyway
$(\Leftarrow)$ Let $x \in X$ and suppose that the zero set nbhds of $x$ form a nbhd base $\mathscr{B}_x$. This means that every nbhd $B \in \mathscr{B}_x$ can be written $B = g^{-1}(0)$ for some continuous $g:X \to \mathbb{R}$.
Not sure what to do from here. Any help for either $(\Leftarrow)$ or $(\Rightarrow)$ is greatly appreciated!
UPDATE: I didn't realize there are varying definitions of completely regular. So let me define it here: A space $X$ is completely regular if whenever $E$ is a closed, nonempty subset of $X$ and $x \not \in E$, then there is a continuous $f:X \to [0,1]$ so that $f(x) = 0$ and $f(E) \subset \{1\}$
Partial answer: $\implies$
Assume that $X$ is completely regular. Let $U$ be an open neigborhood of $x \in X$. Then $E = X \setminus U$ is closed and $x \notin E$. There is a continuous map $f : X \to I$ such that $f(x) = 0$ and $f(E) \subset \{1\}$. The set $N = f^{-1}([0,\frac 1 2]$ is a closed neighborhood of $x$ which is contained in $U$. It remains to show that $N$ is a zero set. Define $$g : X \to I, g(y) = \begin{cases} 0 & y \in N \\ 2f(y)-1 & y \notin N \end{cases}$$ Note that $y \notin N$ means $f(y) \in (\frac 1 2,1]$, thus $2f(y)-1 \in (0,1]$ which means that $g$ is well-defined and has the property $g^{-1}(0) = N$. Clearly $g$ is continuous in all points of the open set $X \setminus N$. Moreover for $0 < \epsilon \le 1$ we get $g^{-1}([0,\epsilon)) = N \cup f^{-1}((\frac 1 2,\frac{\epsilon +1}{2})) = f^{-1}([0,\frac{\epsilon +1}{2}))$ which an open set containing $N$. This proves the continuity of $g$ in all $y \in N$.