Let $A_0,\dots,A_{n-1}$ be sets for some whole $n>0$.
Take $A'_{0, i} = A_i$ and $A'_{1, i} = \bigcup ( \{ A_0, \ldots A_{n - 1} \} \setminus \{A_i\})$ for $i=0,\dots,n-1$.
Prove (or disprove) $$ \bigcup_{i=0}^{n-1} ( A_i \times A_i) = \bigcap_{i=0}^{n-1} \left( (A'_{0, i} \times A'_{0, i}) \cup ( A'_{1, i} \times A'_{1, i}) \right) . $$
Thanks for the clarification. The result is false in general.
Suppose that $n=4$ and we have the following sets: \begin{eqnarray} A_0&=&\{a,x\}\\ A_1&=&\{b,y\}\\ A_2&=&\{c,x\}\\ A_3&=&\{d,y\}. \end{eqnarray}
Then \begin{eqnarray} A'_{1,0}&=&\{b,c,d,x,y\}\\ A'_{1,1}&=&\{a,c,d,x,y\}\\ A'_{1,2}&=&\{a,b,d,x,y\}\\ A'_{1,3}&=&\{a,b,c,x,y\}. \end{eqnarray}
Now, $\{x,y\}\subseteq\bigcap_{i=0}^{3}A'_{1,i}$, so trivially $$\langle x,y\rangle\in \bigcap_{i=0}^{3}\left( (A'_{0,i}\times A'_{0,i})\cup (A'_{1,i}\times A'_{1,i}) \right).$$
However, no single $A_{i}$ has both $x$ and $y$ as an element, so $$\langle x,y\rangle \not\in\bigcup_{i=0}^{3}(A_{i}\times A_{i}).$$
Thus $$\bigcup_{i=0}^{3}(A_{i}\times A_{i})\not\supseteq\bigcap_{i=0}^{3}\left( (A'_{0,i}\times A'_{0,i})\cup (A'_{1,i}\times A'_{1,i}) \right).$$
For what it's worth, the reverse inclusion is fine. It suffices to show that
$$(A_{i}\times A_{i})\subseteq(A'_{0,j}\times A'_{0,j})\cup (A'_{1,j}\times A'_{1,j})$$ for an arbitrary $i,j$. This is trivial for $i=j$ since $A_i=A'_{0,i}$. For $i\neq j$, it follows from the fact that $A_{i}\subseteq A'_{1,j}$.