Well, from your previous question and its excellent answer by Scott, you know what continuously symmetrizable is. In particular we have a $d: X \times X \rightarrow \mathbb{R}$, where $d$ is continuous and $d(x,y) = 0$ iff $x = y$, and this means that $\Delta_X = \{ (x,x) \mid x \in X \}$, the diagonal of $X$ in $X \times X$, can be seen as $d^{-1}[\{0\}]$, and this means by definition that $\Delta_X$ is a zeroset in $X \times X$.
A subset $A \subset Y$, $Y$ a topological space, is called a regular $G_\delta$ iff $A = \cap_{n \in \mathbb{N}} \overline{O_n}$, where the $O_n$ are open sets containing $A$. So in particular such a set is closed (intersection of closed sets) and a $G_\delta$ (as $A = \cap_{n \in \mathbb{N}} O_n$ as well).
Now, it's not too hard to see that a zero-set is a regular $G_\delta$: in the above notation, let $f: A \rightarrow \mathbb{R}$ is continuous and $A = f^{-1}[\{0\}]$. Then define $O_n = f^{-1}[(-\frac{1}{n}, \frac{1}{n})]$ which are open supersets of $A$, and $\overline{O_n} \subset f^{-1}[[-\frac{1}{n}, \frac{1}{n}]]$, and so their intersection equals exactly $A = f^{-1}[\{0\}]$ (it holds in $\mathbb{R}$ and intersection commutes with $f^{-1}[.]$), so $A$ is a regular $G_\delta$. As $\Delta_X$ is a zero-set this shows the second part of the remark in the paper.
All these properties like having a regular $G_\delta$ diagonal, or stronger zero-set diagonal, or even stronger symmetrizable, are examples of so-called generalized metric properties (all metric spaces have them, and they have some sort of metric like structure), which are quite old and well-studied. See this article by Gruenhage for a nice overview by one of the more well-known researchers in this area.