If after confirming the validity of my proof, someone could comment on the possibility of creating a direct proof...that would be awesome. At any rate:
Prove that $\bigcap_{n \lt \omega}f^n[A]=\bigcap_{n\lt \omega}f^n[B]$ using the following information:
i. $B \subseteq A$
ii. $f:A \xrightarrow{1-1} B$
iii. $f^n$ is simply saying iterate $f$ $n$ times (and $f^0$ is the identity function $I_A$)
iv. $A \supseteq B \supseteq f^1(A) \supseteq f^1 (B) \supseteq f^2(A) \supseteq f^2(B) \supseteq...$
The proof to iv. can be found here (as well as a more formal definition of $f^n$): If $B \subseteq A$ and $f:A\overset{1-1}\longrightarrow B$ then $A \supseteq B \supseteq f^1(A) \supseteq f^1 (B) \supseteq f^2(A) \supseteq...$
Importantly, iv. amounts to proving the following two induction claims, which will be used in this proof:
Statement 1. $\forall n \in \omega\big( f^{n+1}[A] \subseteq f^n[B]\big)$
Statement 2. $\forall n \in \omega\big( f^n[B] \subseteq f^n[A]\big)$
Proceed by contradiction: assume that $\bigcap_{n \lt \omega}f^n[A]\neq \bigcap_{n\lt \omega}f^n[B]$ and derive a contradiction.
I have broken this into two cases (based on what is meant by "not equal"):
$ \Big [\bigcap_{n \lt \omega}f^n[A]\neq \bigcap_{n\lt \omega}f^n[B] \Big] \rightarrow \Big [\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B] \ \lor \ \bigcap_{n\lt \omega}f^n[B] \nsubseteq \bigcap_{n \lt \omega}f^n[A] \Big]$
Case One: $\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B]$
Case Two: $\bigcap_{n\lt \omega}f^n[B] \nsubseteq \bigcap_{n \lt \omega}f^n[A]$
Case One:
$\Big [\bigcap_{n \lt \omega}f^n[A]\nsubseteq \bigcap_{n\lt \omega}f^n[B] \Big ] \rightarrow \exists x \Big( x \in \bigcap_{n \lt \omega}f^n[A] \land x \notin \bigcap_{n\lt \omega}f^n[B] \Big ) $ Call one such instance $x'$.
$x' \in \bigcap_{n \lt \omega}f^n[A] \rightarrow \forall n \in \omega \Big (x' \in f^n[A] \Big)$ $\dagger$
$x' \notin \bigcap_{n\lt \omega}f^n[B] \rightarrow \exists n \in \omega \Big (x' \notin f^n[B] \Big )$ Call one such instance $n'$.
So we have $x' \notin f^{n'}[B]$
However, we proved earlier through induction (Statement 1) that $\forall n \in \omega\big( f^{n+1}[A] \subseteq f^n[B]\big)$.
Consider when $n=n'$: we then have $f^{n'+1}[A] \subseteq f^{n'}[B]$
By our earlier universal claim $\dagger$, we know that $x' \in f^{n'+1}[A]$
But $\Big[x' \in f^{n'+1}[A] \land f^{n'+1}[A] \subseteq f^{n'}[B] \Big ] \rightarrow x' \in f^{n'}[B]$.
A contradiction
Case Two works similarly to Case One, except we will use Statement 2 instead of Statement 1. With both claims leading to contradiction, we must conclude that $\bigcap_{n \lt \omega}f^n[A]=\bigcap_{n\lt \omega}f^n[B]$.
I will include the proof for completeness.
Case Two:
$\Big [\bigcap_{n \lt \omega}f^n[B]\nsubseteq \bigcap_{n\lt \omega}f^n[A] \Big ] \rightarrow \exists x \Big( x \in \bigcap_{n \lt \omega}f^n[B] \land x \notin \bigcap_{n\lt \omega}f^n[A] \Big ) $ Call one such instance $x'$.
$x' \in \bigcap_{n \lt \omega}f^n[B] \rightarrow \forall n \in \omega \Big (x' \in f^n[B] \Big)$ $\dagger\dagger$
$x' \notin \bigcap_{n\lt \omega}f^n[A] \rightarrow \exists n \in \omega \Big (x' \notin f^n[A] \Big )$ Call one such instance $n'$.
So we have $x' \notin f^{n'}[A]$
However, we proved earlier through induction (Statement 2) that $\forall n \in \omega\big( f^n[B] \subseteq f^n[A]\big)$
Consider when $n=n'$: we then have $f^{n'}(B) \subseteq f^{n'}(A)$
By our earlier universal claim $\dagger\dagger$, we know that $x' \in f^{n'}[B]$
But $\Big[x' \in f^{n'}[B] \land f^{n'}[B] \subseteq f^{n'}[A] \Big ] \rightarrow x' \in f^{n'}[A]$.
A contradiction $\square$
By straightforward induction, $$ f^{n+1}[A]\subseteq f^n[B]\subseteq f^n[A]$$ hence $$ \bigcap_{n<\omega}f^n[A]\subseteq\bigcap_{0\ne n<\omega}f^n[A]=\bigcap_{n<\omega}f^{n+1}[A] \subseteq \bigcap_{n<\omega}f^{n}[B]\subseteq\bigcap_{n<\omega}f^{n}[A].$$