How to prove this version of the Cantor-Schroder-Bernstein theorem? My text states the Cantor-Schroder-Bernstein theorem as follows:

Suppose that $X$ and $Y$ are non-empty sets such that $|X|>|Y|$. Then, any function $f:X\rightarrow Y$ is not an injection, i.e. there exists distinct elements $x_1$ and $x_2\in X$ such that $f(x_1)=f(x_2)$

My first question is:
This version of the theorem is different than what I have seen in other texts, which typically takes this form.
How is the theorem stated in my text equivalent to the version stated in the Proofwiki? This looks more like the pigeonhole principle to me (without requiring that $X$ and $Y$ be finite sets).
I'd like some help on proving this theorem as stated in my text. What I've tried:
I will prove the contrapositive of the theorem, namely:

For non empty sets $X$ and $Y$, if $f:X\rightarrow Y$ is an injection, then $|X|\le |Y|$

So, if $f$ is an injection, then $f$ may or may not be a surjection.
If $f$ is also a surjection, then we have a bijection $X\rightarrow Y$, which means each $X$ and $Y$ can be paired together. Hence, both $X$ and $Y$ have the same cardinality.
if $f$ is not surjective, then this means $\exists y \in Y,\forall x\in X,f(x)\ne y$. This means that there is at least one element in $y$ that can't be paired with an element in $X$. Hence we conclude that the cardinality of $Y$ is larger that the cardinality of $X$.
Therefore, $|X|\le |Y|$ as required.
Is my proof acceptable? I feel a bit uneasy stating that $X$ and $Y$ can or can not be paired, but given the information in the theorem, I don't know of a more precise way to write this proof.
EDIT:
Asaf Karagila's comment made me reread my text and found the following definition of comparing the cardinalities of two sets:

Two sets $X$ and $Y$ have the same cardinality, written $|X|=|Y|$, if there are equipotent, i.e. there is a bijection $X\rightarrow$ Y.
If there is an injection $X\rightarrow Y$, then we write $|X|\le |Y|$. We write $|X|<|Y|$ to mean that $|X|\le |Y|$ and $|X|\ne|Y|$ and say that $X$ has smaller cardinality than $Y$.

So, in light of the above definition, the contrapositive of the theorem is true because the definition above makes it true. Is that all there is to it?
 A: The answer depends on what's already been proved. But I would say that this is not an acceptable proof. 
Although the negation of $|X| > |Y|$ does indeed turn out to be equivalent  to $|X| \leq |Y|$, that fact is no simpler than what you're asked to prove. So your statement of the contrapositive uses a fact that is probably not permissible. For example, the fact you're using can only be proved with the Axiom of Choice, whereas the Cantor-Bernstein Theorem is valid without it. Briefly: the Cantor-Bernstein Theorem says that $\leq$ is an order relation among cardinalities. The axiom of choice can additionally be used to prove that the order relation is total, but why should you assume this?
Your theorem is equivalent to Cantor-Bernstein for the following reason. Note that $|X| < |Y|$ means, by definition, that $|X| \leq |Y|$ but $|X| \ne |Y|$.
The original Cantor-Bernstein is: If $|X| \leq |Y|$ and $|Y| \leq |X|$, then $|X| = |Y|$.
Your theorem is: If $|Y| < |X|$, then we cannot have $|X| \leq |Y|$.
These two statements are equivalent as a matter of pure logic. 
The first is of the form: (A and B) implies C. 
The second is of the form: (B and not C) implies not A.
Both can be thought of as equivalent to not(A and B and not C).
