Does this proof of the Schröder–Bernstein theorem work? $ f : A \mapsto B $ and $ g : B \mapsto A $ with both functions being injective. Prove that there exists a bijective function $ h $ on $A$ onto $B$.
Suppose $ \exists A_1 $ such that  $ A_1  \subseteq A $ and $ g[ B - f [ A_1 ] ] = A - A_1 $ , in which case we can define a $ h $ such that $ h : A \mapsto B $ where $ x \in A_1 \implies h(x) = f(x) $ and $ x \in A - A_1 \implies h(x) = g^{-1} (x) $ with $h$ being bijective.
Constructing the h is dependent an making an  $ A_1 $ which satisfies $ g[ B - f [ A_1 ] ] = A - A_1 $ . Let $ A_1 = \bigcap \{ A_0 | A - g [B] \subseteq A_0 $ and $ g [ f[ A_0 ] ] \subseteq A_0 \} $.
By defining $ A_1 $ in this way, is $h$ a bijective function? Does this $A_1$ satisfy $ g[ B - f [ A_1 ] ] = A - A_1 $ ?
Here are some thoughts I have on the definition of $ A_1 $ proposed.
Wanting the $ A_1 $ to satisfy the property $ g [ f[ A_1 ] ] \subseteq A_1 $ seems obvious as a direct consequence of injectivity from the statement $ g[ B - f [ A_1 ] ] = A - A_1 $.
The property $ A - g [B] \subseteq A_1 $ probably stems from the fact that if every  element in $A$ that is not in the range of  $g$ is not inside of $ A_1 $ then we could end up having a blindspot in constructing $ h $ to be bijective since we're using $h(x)= f(x)$ for $ x\in A_1 $
I'm slightly confused about how to prove things when the big intersection $ \bigcap $ sign involved.
 A: To find such a set $A_1$, we do not need to find the intersection of all sets $\{ A_0 \mid A-g[B] \subset A_0 \, \, \& \, \, g[f[A_0]] \subset A_0 \}$; in fact, each of such sets can be considered as a good candidate for the set $A_1$. In the discussion below, we construct one of such sets and see that it satisfies the requirement of the proof.
Let us review the idea of the proof of the theorem. Let $A_1$ be a subset of $A$, $f[A_1]=B_1 \subset B$, $B_2=B-B_1$, and $g[B_2]=A_2 \subset A$. If we have some $A_1$ such that $A_1 \cup A_2 =A$, then we have the following bijective mapping between the sets $A$ and $B$:$$h(x)=\begin{cases}f(x) & \text{if }x \in A_1; \\ g^{-1}(x) & \text{if } x \in A_2 \end{cases}.$$Thus, our original problem is reduced to finding the set $A_1$. So, let us find it.
According to the description of the set $A_1$, it must contain the elements of the set $A$ which are not in the range of the function $g$, that is, $A-g[B] \subset A_1$. So one may conclude that the set $C_0=A-g[B]$ is a good candidate for the set $A_1$. But, the set $C_0$ does not satisfy our required condition because$$B_1=f[C_0], \quad B_1 \cap B_2=\varnothing \\ \Rightarrow \quad g[f[C_0]] \cap g[B_2]= \varnothing \quad \Rightarrow \quad C_0 \cup A_2 = A - g[f[C_0]]$$(Please note that in the above argument we have used the fact that $g$ is one-to-one and $g[B_2]=A_2$).
So, we need to add the elements of the set $C_1=g[f[C_0]]$ to the set $C_0$. So one may conclude that the set $C_0 \cup C_1$ is a good candidate for the set $A_1$. But, the set $C_0 \cup C_1$ does not satisfy our required condition because$$B_1=f[C_0 \cup C_1], \quad B_1 \cap B_2=\varnothing , \quad C_1 \cap g[f[C_1]]= \varnothing \\ \Rightarrow \quad g[f[C_0 \cup C_1]] \cap g[B_2]= \varnothing \\ \Rightarrow \quad (C_0 \cup C_1) \cup A_2 = A - g[f[C_1]]$$(Please note that in the above argument we have used the fact that $g$ and $f$ are one-to-one and $g[B_2]=A_2$).
$$\vdots \qquad \vdots \qquad \vdots$$
So, we need to add the elements of the set $C_n=g[f[C_{n-1}]]$ to the set $\bigcup_{i=0}^{n-1}C_i$. So one may conclude that the set $\bigcup_{i=0}^n C_i$ is a good candidate for the set $A_1$. But, the set $\bigcup_{i=0}^n C_i$ does not satisfy our required condition because$$B_1=f \left [\bigcup_{i=0}^n C_i \right ], \quad B_1 \cap B_2=\varnothing , \quad C_n \cap g[f[C_n]]= \varnothing \\ \Rightarrow \quad g \left [f \left [\bigcup_{i=0}^n C_i \right ] \right ] \cap g[B_2]= \varnothing \\ \Rightarrow \quad \left (\bigcup_{i=0}^n C_i \right ) \cup A_2 = A - g[f[C_n]]$$(Please note that in the above argument we have used the fact that $g$ and $f$ are one-to-one and $g[B_2]=A_2$).
$$\vdots \qquad \vdots \qquad \vdots$$
Hence, we can conclude that the set $\bigcup_{n=0}^{\infty } C_n$ satisfies our required condition because$$g \left [f \left [ \bigcup_{n=0}^{\infty } C_n \right ] \right ] \subset \bigcup_{n=0}^{\infty } C_n \\ \Rightarrow \quad \left ( \bigcup_{n=0}^{\infty }C_n \right ) \cup g \left [B- f \left [ \bigcup_{n=0}^{\infty } C_n \right ] \right ]=A.$$Thus,$$A_1=\bigcup_{n=0}^{\infty }C_n, \quad \text{ where } \, C_0=A-g[B], \quad C_n=g[f[C_{n-1}]].$$

Addendum
Please note that the idea of the foregoing proof is nothing but the one we see in the König's proof of the Cantor-Schroeder-Bernstein Theorem. To see that fact, let us first review the König's proof.
König's Proof
Let $a \in A$, then $a \in A-g[B]$ or $a \in g[B]$. In the first case $a$ has no inverse image under $g$ while in the second case it has an inverse image $g^{-1}(x)$. In the second case, either $g^{-1}(a) \in B- f[A]$, in which case $g^{-1}(a)$ has no inverse image under $f^{-1}$, or $g^{-1}(a) \in f[A]$. Again, in the latter cases there exists a unique inverse image $f^{-1}(g^{-1}(a))$, ... . We call the elements $g^{-1}(a)$, $f^{-1}(g^{-1}(a))$, ... the ancestors of $a$. By similar analogy, such a description also holds for an arbitrary element $b \in B$.
Now, we can partition $A$ into three subsets:

*

*$A_e$: the set of all elements of $A$ having an even number of ancestors,


*$A_o$: the set of all elements of $A$ having an odd number of ancestors,


*$A_i$: the set of all elements of $A$ having an infinite number of ancestors.
Similarly, we can partition $B$ into such subsets. It is clear that if $x\in A_e$ then $f(x) \in B_o$, if $x\in A_o$ then $g^{-1}(x) \in B_e$, and if $x\in A_i$ then $f(x) \in B_i$. So, we can easily conclude that the restricted functions $f:A_e \to B_o$, $g^{-1}: A_o \to B_e$, and $f:A_i \to B_i$ are bijective.
Here is the idea of the proof. Since $f$ and $g$ are injective functions into $B$ and $A$, respectively, one may want to map any element of $A$ into $B$ by $f$ and cover the remaining elements of $B$, i.e., $B-f[A]$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)=\begin{cases}f(x) & \text{if } x \in A-g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[B-f[A]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).
But, the obtained mapping is not onto $B$ because its range lacks the subset $f[g[B-f[A]]]$ (that is, the values of $f$ of the $1$-ancestor elements of $A$). So, one may want to cover such a subset of $B$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)= \begin{cases}f(x) & \text{if }x \in A-(g[B-f[A]] \cup g[f[g[B-f[A]]]]); \\ g^{-1}(x) & \text{if }x \in g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[f[g[B-f[A]]]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]] \cup g[f[g[B-f[A]]]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).
But, the obtained mapping is not onto $B$ because its range lacks the subset $f[g[f[g[B-f[A]]]]]$ (that is, the values of $f$ of the $3$-ancestor elements of $A$).
$$\vdots \qquad \vdots \qquad \vdots$$
But, the obtained mapping is not onto $B$ because its range lacks the subset $(f \circ g)^n[B-f[A]]$ (that is, the values of $f$ of the $2n-1$-ancestor elements of $A$). So, one may want to cover such a subset of $B$ by $g^{-1}$ so that the following injective mapping is obtained:$$h(x)= \begin{cases}f(x) & \text{if }x \in A-(g[B-f[A]] \cup g[f[g[B-f[A]]]] \cup \cdots \cup (g \circ f)^n[g[B-f[A]] ); \\ g^{-1}(x) & \text{if }x \in g[B-f[A]]; \\ g^{-1}(x) & \text{if } x \in g[f[g[B-f[A]]]]; \\ \vdots & \vdots \\ g^{-1}(x) & \text{if } x \in (g \circ f)^n[g[B-f[A]] \end{cases}$$(Please note that we have to remove the set $g[B-f[A]] \cup g[f[g[B-f[A]]]] \cup \cdots \cup (g \circ f)^n[g[B-f[A]]$ from the set $A$ in the domain of the first piece so that $h$ becomes a function).
But, the obtained mapping is not onto $B$ because its range lacks the subset $(f \circ g)^{n+1}[B-f[A]]$ (that is, the values of $f$ of the $2n+1$-ancestor elements of $A$).
$$\vdots \qquad \vdots \qquad \vdots$$
This pattern motivates us to define mapping $h$ as follows.$$h(x)=\begin{cases}f(x) & \text{if } x \in A_e \cup A_i; \\ g^{-1}(x) & \text{if } x \in A_o \end{cases}$$(Please note that $\bigcup_{n=0}^{\infty }(g \circ f)^n[g[B-f[A]]=A_o$ and $A-\bigcup_{n=0}^{\infty }(g \circ f)^n[g[B-f[A]]=A_e \cup A_i$).
Since $h$ maps the set $A=A_e \cup A_i \cup A_o$ onto the set $B=B_o \cup B_i \cup B_e$ (as already explained) and the sets $A_e \cup A_i$ and $A_o$ are disjoint, we conclude that the mapping $h$ is bijective. $\square$
Now, comparing the mentioned proofs with each other, we can easily see that the $C_n$'s constructed in our proof are nothing but $A_{2n}$'s in the König's proof, that is, the sets of elements of $A$ having $2n$ ancestors. So, it is clear that$$\bigcup_{n=0}^{\infty }C_n=A_e \cup A_i.$$
You can also find other approaches to prove the Cantor-Schroeder-Bernstein theorem in this post.
