Let $R$ be a commutative ring with identity and suppose that $J = \bigcup I_\alpha$ is a chain of radical ideals such that $a \in R$ is a zero divisor in every $R/I_\alpha$. Is $a$ necessarily a zero divisor in $R/J$?
I don't expect this to be true, but I have not been able to think of any counterexamples, and this would be a useful result.
(If conversely $a$ is a zero divisor in $R/J$, then trivially it is a zero divisor in $R/I_\alpha$ for all sufficiently big $I_\alpha$.)
If this held, then Zorn's lemma would guarantee a factor ring $S=R/J$ of $R$ with an element $a \in S$ such that $Ann(a) \subset Ann(b)$ whenever $b \notin Ann(a)$ (follows because $a$ is by construction regular in $S / Ann(b)$). One can then check that the diameter of the zero divisor graph of $S$ would be at most $2$, which would imply that either the zero divisors of $S$ form an ideal or that $S$ has exactly $2$ minimal primes (see e.g. Theorem 2.1 here)
Let $k$ be a field and $R=k[x,y_1,y_2,\dots]/(xy_1,xy_2,\dots)$ and $I_n=(y_1,\dots,y_n)$. Then $x$ is a zero divisor in $R/I_n$ for any $n$ but it is not a zero divisor in $R/\bigcup I_n\cong k[x]$.