Recently I constructed a special topology space which is modifying the example 2.17 of Arhangel'skii as follows :
The space is $ Z=X_0\cup X_1\cup X$, where $X_0=\Bbb R\times\{0\}$, $X_1=\Bbb R\times\{-1\}$, and $X=\Bbb R\times(0,\to)$. For $x=\langle a,0\rangle\in X_0$ let $x'=\langle a,-1\rangle\in X_1$. For $n\in\Bbb Z^+$ and $x=\langle a,0\rangle\in X_0$ let
$V_n(x)=\{x\}\cup$ a triangle which has the sides adjacent to the vertex $x$ of equal length and an angle at $x$ of measure $30^0$ with height equal to $\frac1n$ and height slope equal to $1$.
$V_n(x')=\{x'\}\cup$ a triangle which has the sides adjacent to the vertex $x$ of equal length and an angle at $x$ of measure $30^0$ with height equal to $\frac1n$ and height slope equal to $-1$.
The topology on $Z$ obtained by this: The topology $\tau$ on $X$ is the smallest topology obtained by $\tau_1\cup \tau_2$, where $\tau_1$ is Euclidean topology on $X$ and $\tau_2$ is the topology of countable complements on $X$ (Much more details see 63# of counterexample in topology); and we take the families $\{V_n(x):n\in\Bbb Z^+\}$ and $\{V_n(x'):n\in\Bbb Z^+\}$ as local bases at $x\in X_0$ and $ x' \in X_1$, respectively.
Does this space have a zeroset diagonal? Any idea will be appreciated:)