This might even be a notational nuisance, but here it goes.
Let $\kappa$ be a cardinal, $X_\alpha\subseteq\kappa$ for all $\alpha<\kappa$. As you know, the diagonal intersection of $\{X_\alpha\}_\alpha$ is the set $$ \triangle_\alpha X_\alpha := \{\xi\in\kappa : \xi\in \textstyle\bigcap_{\alpha<\xi} X_\alpha\}. $$ Now, does $0$ belong to this set?
It should be the case when $0\in\bigcap_{\alpha<0} X_\alpha = \bigcap \emptyset$, which is undefined, or $\emptyset$ by convention (in some books). In the last case, I would conclude that $0\notin\triangle_\alpha X_\alpha$; but there are comments both in Kunen and in Jech that imply the contrary. For instance, Jech writes (after the previous definition) $$ \triangle_\alpha X_\alpha=\textstyle\bigcap_\alpha(X_\alpha\cup\{\xi:\xi\leq\alpha\}). $$ Actually, if you define $\bigcap$ as a function from classes to classes, $\bigcap \emptyset = V$ makes perfect sense, but even there Jech takes care of restricting its domain to nonempty classes.
A meta-question: Is there some context where this matters?
All the definitions given seem to agree that $0$ is always in the diagonal intersection. The sets
and
are the same set; in particular, $0$ is in the latter, since for every $\alpha\in\kappa$ we have $\alpha\ge0$, so $0\in\{\xi: \xi\le\alpha\}$ and hence is in $X_\alpha\cup\{\xi: \xi\le\alpha\}$.
I don't see any definition which makes $0$ not an element of the diagonal intersection, and I don't see any remark in Jech or Kunen where this is contradicted.