I am attempting to prove the following:
For each $A, B \in \wp(Y)$ with $A \subset B$ , prove that $f^\leftarrow(A) \subset f^\leftarrow(B)$
NOTE: It should be noted that $f^\leftarrow$ is the powerset operator $f^\leftarrow: \wp(Y) \to \wp(X)$ for some function $f:X \to Y$ which is NOT to be confused with $f^{-1}$, which is $f$'s inverse (if one even exists). $f^\leftarrow(D)$ returns the preimage of $D$ but does not assert the existence of an inverse function $f^{-1}$.
My sketch of the proof looks like the following:
We would first start out by noting that since $A \subset B$, $\forall x \in A, x \in B$. Then we would begin to think of $A$ and $B$ as the images of some other sets $A_X, B_X \subset X$, i.e. we would say
$$A =: f^\to(A_X) \hspace{1cm} B =: \ f^\to(B_X)$$
$\uparrow$ This final assertion is the trickiest one because we do not know that $f$ is surjective. I suspect I have made too many jumps in my thinking. Any ideas on where to go from here?
I would suggest approaching the proof in a more structured fashion. A proof is a very precise mathematical object, however we always write out a proof in a mixture of English (or your preferred natural language) and formal mathematics. However, it is worthwhile to remember that very often a proof starts out with a process of being precise about what exactly, syntactically, it is that we wish to prove. These steps do not involve thinking. Only reciting definitions and copying text around. In this case, the proofs evolves as follows. Let $A\subseteq B \subseteq Y$ be given. We wish to show that $f^\leftarrow (A) \subseteq f^\leftarrow(B)$. The meaning of that which is given is: $y\in A\implies y\in B$, so that is known and can be used in our proof. The meaning of that which we like to prove is $f(x)\in A\implies f(x)\in B$. So, let us assume that for some $x\in X$ it is the case that $f(x)\in A$. Our goal is to show that $f(x)\in B$. Up to this point the steps are mechanical. It is time to use that $y\in A\implies y\in B$. Let $y=f(x)$. Then $y\in A$, and thus $y\in B$, so $f(x)\in B$. QED.
The point of this answer is to demonstrate how 90% of the proof involves mechanical steps following the rules of what does it means to prove a certain claim.