Need help with proof in elementary set theory At the risk of stretching the patience of readers here, I will try this once more. (See: Need help with proof about Dedekind-infinite sets for my initial debacle.)
Suppose we have


*

*$C \subseteq B \subseteq A$

*Bijection $f:A\rightarrow C$

*$\forall x (x\in A \wedge x\notin C \rightarrow \neg\exists y (y\in A \wedge f(y)=x))$  (Edit: Turns out to be redundant)


Can we prove there exists bijection $g:B\rightarrow C$?
Hopefully without invoking the convoluted machinery of the Cantor-Bernstein-Schroder proof?
 A: Condition 3 follows from 2, as the codomain of $f$ is $C$. 
Now, if you have a proof that 1 and 2 imply that there is a bijection between $B$ and $C$, then you would have a proof of Cantor-Bernstein. And no one has been able to prove Cantor-Bernstein without the "convoluted machinery".
A: Let me try to describe the mental picture that I mentioned in my comment to Martin Argerami's answer.  (Unfortunately, I have no idea how to produce an actual picture here, so I'll have to settle for a description.  I assure you the picture can be drawn more quickly than the description can be read.)  Imagine $A$ as a long horizontal rectangle; imagine $A-B$ as a small sub-rectangle $A_1$ at the right end; and therefore imagine $B$ as a long (but not quite so long as $A$) subrectangle of $A$, extending from the left end of $A$ to well beyond the middle.  Imagine $B-C$ as a small rectangle $B_1$ at the left end of $B$; so $C$ is a (still rather long) rectangle occupying the middle of $A$.  The given bijection $f$ maps $A$ into this sub-rectangle.  Let $A_2=f[A_1]$ and imagine it as occupying the right end of $C$, adjacent to $A_1$; similarly, imagine $B_2=f[B_1]$ as occupying the left end of $C$, adjacent to $B_1$; so $f[C]$ is the middle part of $C$.  $A_2$ and $B_2$, being part of $C$, are mapped by $f$ into this middle part $f[C]$.  Imagine $A_3=f[A_2]$ at the right end of $f[C]$, $B_3=f[B_2]$ at the left end, and $f^2[C]$ (by which I mean $f[f[C]]$) in the middle.  Continue in this way with $A_n=f[A_{n-1}]$ at the right and $B_n=f[B_{n-1}]$ at the left end of $f^{n-2}[C]$, with $f^{n-1}[C]$ in the middle.  
So we have, starting at the right end, the sequence of sets $A_1,A_2,\dots$, each mapped to the next by $f$; similarly, starting at the left end, we have $B_1,B_2,\dots$ each mapped to the next by $f$.  In the middle, there might be some points in $\bigcap_{n\in\mathbb N}f^n[C]$.
Now we can biject $B$ to $C$ by applying $f$ to everything in the $B_n$'s (shifting each $B_n$ one step to the right) while leaving everything else ($A_n$'s and $\bigcap_{n\in\mathbb N}f^n[C]$) fixed.
A: I don't know what's so convoluted about the proof of the Cantor-Bernstein theorem. It's a corollary of Knaster's fixed-point theorem:
For any function $\varphi:\mathcal P(C)\rightarrow\mathcal P(C)$ such that $X\subseteq Y\subseteq C\Rightarrow\varphi(X)\subseteq\varphi(Y)$, there is a set $X\subseteq C$ such that $\varphi(X)=X$.
The proof of Knaster's theorem can't require all that much "convoluted machinery", because it was a Putnam problem in 1956 or 1957. (It must have been the easiest problem on the Putnam that year, because it's the one I solved, not on the exam but by the next day.) To prove it, just define $X=\bigcup\mathcal S$ where $\mathcal S=\bigcup\{Z:Z\subseteq\varphi(Z)\}$, and show that $\varphi(X)=X$.
To see this, note that if $Z\in\mathcal S$ then $Z\subseteq X$ and so $Z\subseteq\varphi(Z)\subseteq\varphi(X)$. Since the elements of $\mathcal S$ are subsets of $\varphi(X)$, it follows that $X=\bigcup\mathcal S\subseteq\varphi(X)$ and so $\varphi(X)\subseteq\varphi(\varphi(X))$, which means that $\varphi(X)\in\mathcal S$, whence $\varphi(X)\subseteq X$.
Now suppose that $C\subseteq B\subseteq A$ and that $f:A\rightarrow C$ is an injection. Define $\varphi:\mathcal P(C)\rightarrow\mathcal P(C)$ by setting $\varphi(X)=C\setminus f(B\setminus X)$. Seeing as $X\subseteq Y$ implies $B\setminus X\supseteq B\setminus Y$ implies $f(B\setminus X)\supseteq f(B\setminus Y)$ implies  $C\setminus f(B\setminus X)\subseteq C\setminus f(B\setminus Y)$ implies $\varphi(X)\subseteq\varphi(Y)$, it follows by Knaster's theorem that $\varphi(X)=X$ for some $X\subseteq C$, i.e., $C\setminus f(B\setminus X)=X$, or in other words, $f(B\setminus X)=C\setminus X$. Thus we get a bijection $g:B\rightarrow C$ by defining $g(x)=x$ for $x\in X$ and $g(x)=f(x)$ for $x\in B\setminus X$.
