Is my proof from elementary Set Theory correct? Let $\{A_i : i\in I\}$ and $\{B_i : i\in I\}$ be indexed families of sets.
Show that:
$$(\forall i\in I)(A_i \subseteq B_i) \implies \bigcup_{i \in I} A_i \subseteq \bigcup_{i \in I} B_i$$
My proof:
Let's assume that $(\forall i\in I)(A_i \subseteq B_i)$. If $x \in A_i$ for some $i \in I$, then It results from the assumptions that also $x \in B_i$, therefore $(\exists i \in I)(x \in A_i) \implies (\exists i \in  I)(x \in B_i)$, what is equivalent to $x\in\bigcup_{i \in I} A_i \implies x \in \bigcup_{i \in I} B_i$.
$$(x\in\bigcup_{i \in I} A_i \implies x \in \bigcup_{i \in I} B_i) \equiv (\bigcup_{i \in I} A_i \subseteq \bigcup_{i \in I} B_i)$$QED.
 A: All in all I think it's fine. The full argumentation chain is:
$x \in \cup_{i \in I} A_i \Rightarrow \exists k \in I : x \in A_k$.
$ x \in A_k \Rightarrow  x \in B_k \Rightarrow x \in \cup_{i \in I} B_i$. 
A: I think it is better that you want to start picking an $x \in \bigcup_{i \in I}A_i$ and then show it belongs to $\bigcup_{i \in I}B_i$, instead of picking an $x \in A_i$. Then, if you start by letting $x \in \bigcup_{i \in I}A_i$, you can use the assumptions to get to $x \in \bigcup_{i \in I}B_i$. It will look like:
Proof:  Let $x \in \bigcup_{i \in I}A_i$, then we have that $x \in A_i$ for some $i$. But since $A_i \subset B_i$ for all $i \in I$, we must have that $x \in B_i$ too for some $i$. As this works for every element, we have that our $x$, where we started with $x \in \bigcup_{i \in I}A_i$, must also belong to a $B_i$, and hence we have that for all $x:$ $\bigcup_{i \in I}A_i \subset \bigcup_{i \in I}B_i$
I would suggest that in these kind of proofs it is always better to start with picking an element of the first set and then show, by using the conditions, it belongs to the second set. 
