For the set of all Injective functions $f: X \to Y$, does $|X|=1$ I was given the following problem worded exactly:

"Let $X,Y \neq \emptyset$ and suppose that every element of $F(X,Y)$ is injective. Show that $|X|=1.$"

and the set  $F(X,Y) := \{f\ | f:X \to Y \} $
I can't prove this statement holds, in fact I think it is false. I came up with the following argument:
Suppose that every element of $F(X,Y)$ is injective, and modifying the definition above we get the set $$F(X,Y) := \{f\ | f:X \to Y \ \wedge f(x_1)=f(x_2)\ \supset \ \ x_1=x_2\} $$
Take set $X=\{x_1,x_2,\dots,x_n\}$ and set $Y=\{y_1,y_2,\dots,y_m\}$ $n\le m$
and we can construct a counterexample where $f_1$ is injective and maps $x_i\to y_i$ for all $i=1,...,n$.
Or similarly lets say $f_2$ is injective and maps $x_i\to y_{2i}$ for all $i=1,...,n$ where $y_{2i}\le m$. 
This would mean $|X|\neq1$. 
However I can see that if $|X|=1$ and $Y$ was nonempty then necessarily  $f:X \to Y$ would be injective. 
Can you help explain if I'm wrong, how I am wrong and what the proof would be that shows the statement above holds? Thanks.
 A: Your error is that the existence of some injective functions $f\colon X \to Y$ doesn't disprove the claim that "if all functions $f \colon X \to Y$ are injective then $|X| = 1$." Indeed, by assuming $|X| \le |Y|$, there are injective functions from $X$ to $Y$, yet the claim is still true.
Instead of using contradiction, to prove that 
$$\text{If every function in $F(X, Y)$ is injective, then $|X| = 1$}$$ it's much easier to prove the contrapositive: 
$$\text{If $|X| > 1$, then there exists some function $f \in F(X, Y)$ that's not injective}$$
(we're assuming $X, Y$ are nonempty, so the negation of $|X| = 1$, that is $|X| \neq 1$, amounts to $|X| > 1$). 
I think you'll have a much easier time believing this is true, and even proving it. Just pick some generic sets $X, Y$ with $|X| > 1$ and see if you can find a way to build a function that isn't injective. Before long you'll have some ideas about how to prove it for general $X, Y$. 
A: If $a,b\in X$ with $a\ne b$ then for any $c\in Y$ the function $X\times \{c\}$ is not injective. 
Note: You might call $X\times  \{c\}$ the graph of a function. In set theory, a function IS its graph.
