Let $\lambda$ a cardinal and $\delta<\lambda^+$. I want to proof there exists a increasing chain $\{A^i_\delta : i< cf(\lambda)\}\subseteq[\delta\times\delta]^{<\lambda}$ converging to $\delta\times\delta$.
If $\delta<\lambda$ then $|\delta\times\delta|=|\delta|<\lambda$ and let $A^i_\delta=\delta\times\delta$ for all $i< cf(\lambda)$ and it's ok.
If $\lambda\leq\delta<\lambda^+$ then $|\delta|=\lambda$. Choose a (strictly) increasing sequence $\langle\alpha_\xi : \xi< cf(\lambda)\rangle$ so that $\sup(\alpha_\xi)=\lambda$ and let $\phi$ a bijection between $\lambda$ and $\delta\times\delta$. Let $B^i_\delta=\phi(\alpha_i)\subset \delta\times\delta$ for all $i< cf(\lambda)$. Then $|B^i_\delta|<\lambda$. But it is not necessary an increasing sequence so define by induction $A^i_\delta$ as follows : $\begin{align*} A^0_\delta&=B^0_\delta\\ A^1_\delta&=A^0_\delta\cup B^1_\delta\\ &\vdots\\ A^{i+1}_\delta&=A^i_\delta\cup B^{i+1}_\delta&\qquad\text{successor}\\ A^i_\delta&=\bigcup_{j First, $\bigcup_{i