Method for proving $ f^{-1}(G \cup H) = f^{-1}(G)\cup f^{-1}(H) $ Prove that if $ f: A \rightarrow B $ and $ G,H $ are subsets of $ B $, then $ f^{-1}(G \cup H) = f^{-1}(G)\cup f^{-1}(H) $.
My (incorrect) Attempt:  Suppose $ x \in f^{-1}(G\cup H) $.  Then there exists a $ y \in G \cup H $ such that $ f^{-1}(y)=x $.  Since $ y \in G \cup H $, $ y\in G $ or $ y \in H $.  Without loss of generality, suppose $ y \in G $.  Then $ x \in f^{-1}(G) $.  So $ x \in f^{-1}(G) \cup f^{-1}(H) $. 
Assume $ x \in f^{-1}(G) \cup f^{-1}(H) $.  Then $ x\in f^{-1}(G) $ or $ x\in f^{-1}(H) $.  Without loss of generality, suppose that $ x\in f^{-1}(G) $.  So there exists some $ y \in G $ such that $ f^{-1}(y)   = x $.  Since $ y \in G $, $ G \cup H $.  Thus $ x \in f^{-1}(G \cup H) $. QED
It has been a while since I last did proofs for inverse functions, and I recall that my initial way of doing them back in day was wrong.  Does this method work?  Thanks
 A: (This is not an evaluation of your proof, but an alternative one.  For clarity I'll be using a slightly different notation: $\;f^{-1}[Y]\;$ instead of $\;f^{-1}(Y)\;$.)
Let's start at the most complex side, here the right hand side, and see if we can use the definitions to find out which elements $\;x\;$ are in this set:
\begin{align}
& x \in f^{-1}[G] \cup f^{-1}[H] \\
\equiv & \;\;\;\;\;``\text{definition of $\;\cup\;$"} \\
& x \in f^{-1}[G] \;\lor\; x \in f^{-1}[H] \\
\equiv & \;\;\;\;\;``\text{basic property $(0)$, see below, twice"} \\
& f(x) \in G \;\lor\; f(x) \in H \\
\equiv & \;\;\;\;\;``\text{reintroduce $\;\cup\;$ using its definition -- really the only thing we can do"} \\
& f(x) \in G \cup H \\
\equiv & \;\;\;\;\;``\text{reintroduce $\;\cdot^{-1}[\cdot]\;$ using basic property $(0)$"} \\
& x \in f^{-1}[G \cup H] \\
\end{align}
By set extensionality, this proves the original statement.
The basic property of $\;\cdot^{-1}[\cdot]\;$ used here is
$$
(0) \;\;\; x \in f^{-1}[Y] \;\equiv\; f(x) \in Y
$$
