This is my first time attempting to try to prove something pertaining to an indexed collection of sets. Is the proof that follows correct? I feel as if there's a fatal flaw.
Observation 1. Suppose $\{ A_i : i \in I \}$ and $\{ B_i : i \in I \}$ are indexed families of sets, then $\bigcup _{i \in I} (A_i \times B_i) \subseteq \left( \bigcup_{i \in I} A_i \right) \times \left( \bigcup_{i \in I} B_i \right)$.
Proof. Let $(x, y)$ be arbitrary elements of $ \bigcup _{i \in I} (A_i \times B_i)$. Then $x \in \bigcup_{i \in I} A_i$, and we have that $x \in A_i$ for some $i \in I$. Similarly, $y \in \bigcup_{i \in I} B_i$, and we have that $y \in B_i$ for some $i \in I$. So, $(x, y) \in \left( \bigcup_{i \in I} A_i \right) \times \left( \bigcup_{i \in I} B_i \right)$. Since $(x, y)$ were arbitrary, then $\bigcup _{i \in I} (A_i \times B_i) \subseteq \left( \bigcup_{i \in I} A_i \right) \times \left( \bigcup_{i \in I} B_i \right)$.
You're using the definitions in the wrong order. It's not that "$x\in A_i$ for some $i\in I$, and $y\in B_i$ for some $i\in I$", because that language indicates that the $i$ for $x$ and the $i$ for $y$ may differ (the context of the quantifier ends before the next one begins).
But let $X_i$ be $A_i\times B_i$, then you assume that $(x,y)\in\bigcup X_i$, so there is some $i\in I$ such that $(x,y)\in X_i$. In other words, there is some $i\in I$ such that $x\in A_i$ and $y\in B_i$. Here it is clear that this is the same $i$ for both.
Now we can move to the separate unions: $x\in A_i$, therefore $x\in\bigcup_{j\in I}A_j$, and $y\in B_i$ so $y\in\bigcup_{j\in I}B_j$. And so on.