Let $f:X\rightarrow Y$. Let $A$, $A_1$ and $A_2$ be subsets of $X$ and $B$, $B_1$, and $B_2$ be subsets of $Y$. Then, I need to prove that $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$.
I know that by the definition of an inverse image, if we let $x\in f^{-1}(B_1\cup B_2)$, then we know that $f^{-1}(x)\in(B_1\cup B_2)$. From here, I'm not sure how to get to $f^{-1}(B_1)\cup f^{-1}(B_2)$.
On
Let $x \in f^{-1}(B_1 \cup B_2)$. Then there exists $y\in B_1 \cup B_2$ with $f^{-1}(y) = x$. Since $y \in B_1 \cup B_2$ then $y$ is in one of $B_1$ or $B_2$ (or both). Let's assume (without loss of generality) that $y\in B_1$. Then since $f^{-1}(y) = x$ and $y \in B_1$ then $x \in f^{-1}(B_1)$. So $x\in f^{-1}(B_1) \cup f^{-1}(B_2)$. The $x$ that we picked at the beginning was arbitrary so we have that $$ f^{-1}(B_1 \cup B_2) \subseteq f^{-1}(B_1) \cup f^{-1}(B_2). $$
Now you just need to show that $$ f^{-1}(B_1) \cup f^{-1}(B_2) \subseteq f^{-1}(B_1 \cup B_2) $$ and you're done.
From where you started, then $f(x) \in B_1\cup B_2 \Rightarrow x \in f^{-1}(B_1)\cup f^{-1}(B_2)$, conversely if $x \in f^{-1}(B_1) \cup f^{-1}(B_2) \Rightarrow f(x) \in B_1\cup B_2 \Rightarrow x \in f^{-1}(B_1\cup B_2)$, and the equality follows.