Suppose $(X,\tau)$ is a topological space and that $(X^2,\tau_2)$ is the product space. Now define
$\mathscr S\!_\tau=\{W\in\tau_2|\Delta X^2\subseteq W\}$, where $\Delta X^2=\{(x,y)\in X^2|x=y\}$,
and suppose $(x_n)$ is a sequence in $X$ such that
$\forall S\in\!\mathscr S\!_\tau\exists N\in\mathbb N:(n,m>N\implies(x_n,x_m)\in S)$.
Is it then possible to deduce that $(x_n)$ has a clusterpoint in $X$,
$\exists x\in \!X\;\forall \mathcal O\in\tau: (x\in\mathcal O\implies\forall N\in\mathbb N\,\exists n>N: x_n\in\mathcal O)$?
Yes, at least if $X$ is Hausdorff. We can prove the contrapositive: Assume that $(x_n)_n$ does not have any accumulation points, and we will then prove that it is not quasi-Cauchy.
Let $A=\{x_n\mid n\in \mathbb N\}$. Since the sequence has no accumulation points, every subset of $A$ is closed.
In particular, for every $x_n$ the set $B_n=X\setminus (A\setminus \{x_n\})$ is open, and the union of all the $B_n$s is $X$. Now $$ S = \bigcup_n (B_n)^2 $$ is in $\mathscr S$, but $(x_n)_n$ doesn't satisfy the quasi-Cauchy criterion with respect to this $S$. Namely, $(x_i,x_j)\in S$ only if $x_i=x_j$, and if that happens for all $i,j>N$, then $x_n$ is constant from some point, contradicting the assumption that it has no accumulation points.