i am trying to prove $ (\bigcup_\ S) \times (\bigcup_\ T) \subset (\bigcup_\ \{ X \times Y | X \in S, Y \in T\}) $. And this seems straight forward, but for some reason i have a hard time applying the definitions without getting confused. I start by $ (u,v) \in (\bigcup_\ S) \times (\bigcup_\ T); \space u \in \bigcup_\ S \space and \space v \in \bigcup_\ T \\ So \space u \in X, \space X \in S \space and \space v \in Y, Y \in T \\ (u,v) \in \{(x,y) | x \in X, y \in Y, X \in S, Y \in T\} \\ (u,v) \in \{(x,y) \in X \times Y |X \in S, Y \in T\} $.
I am really not sure if this is even correct usage of the definition, i am really confused at this. Is there any method i should consider to keep the definitions straight and apply them correctly? If i understand correctly i want to arrive at something that looks like this.
$ (u,v) \in \{(x,y)| (x,y) \in A, A \in X \times Y\} $
I think it would help to rewrite the problem; the exercise uses a more concise notation to denote the union of all sets in a collection, but that obscures what each of the summands (uniands?) in the union looks like. You can write this more clearly as
$(\bigcup_{X\in S}X)\times (\bigcup_{Y\in T}Y)\subseteq \bigcup_{X\in S, Y\in T}X\times Y$.
Then, as you say, if $(u,v)$ is a term in the set on the left hand side, then $u\in X_0$ for some $X_0\in S$, and $v\in Y_0$ for some $Y_0\in T$. Then you must have $(u,v)\in X_0\times Y_0$, so it's in the set on the right hand side. Your argument is correct, but it can be useful to see exactly what the union being taken is.