Prove $f^{-1}(E\cup F)=f^{-1}(E)\cup f^{-1}(F)$
This is from the textbook we're using in my Analysis class:
$x\in f^{-1}(E \cup F)\leftrightarrow f(x) \in (E\cup F)$
$\leftrightarrow(f(x)\in E) \vee (f(x)\in F)$
$\leftrightarrow (x\in f^{-1}(E)) \vee (x\in f^{-1}(F))$
$\leftrightarrow (x\in f^{-1}(E)) \cup (x\in f^{-1}(F))$
Can you help me understand the reasoning behind this proof? I'm not sure I get it.
Given the function $f:D \rightarrow K$ and $E \subseteq K$, the preimage of a set E is a set defined as $f^{-1}(E) :=\{ x \in D: f(x) \in E\}$. We need to prove that the sets $f^{-1}(E\cup F)$ and $(f^{-1}(E)\cup f^{-1}(F))$ are equal. Two sets are equal iff they contain the same elements. As I understand it, they wanted to prove that if an element $x$ is in $f^{-1}(E \cup F)$ then it must also be in $(f^{-1}(E) \cup f^{-1}(F))$ and vice versa. First they used the definition of the preimage of set, and then the definition of the union of sets. And then they again used the definition of the preimage. I still don't see how the last row of the proof implies that $f^{-1}(E\cup F)=f^{-1}(E)\cup f^{-1}(F)$. They showed it's true for $x$ (though I'm not sure I understand that either) but how then do they know it's true $\forall x$?
$$(x\in f^{-1}(E))\vee(x\in f^{-1}(F))\leftrightarrow x\in f^{-1}(E)\cup f^{-1}(F) $$
it is true for any arbitrary element $x\in f^{-1}(E\cup F)$, i.e.: if $x$ is an element of $ f^{-1}(E\cup F)$, then $x$ is also in $f^{-1}(E)\cup f^{-1}(F)$ and vice versa because there is "$\leftrightarrow$" signs in every step, that means you can go backward and obtain the statement for the other way.
Recall the definition of set equality: $E=F$ if $E\subset F$ and $F\subset E$.
And also the definition of subset relationship: $E\subset F$ if $\forall x\in E, E\in F$
The given answer is basically by definition. To check $\forall x\in E, E\in F$, you let an arbitrary element $x\in E$(In case $E$ is not empty), then you check if $x\in F$.