Proving an intuitively true statement. Let $X \subseteq Y$ and $X\neq Y$ Also let  $f: Y → X$ define a bijection. Prove that $Y$ is infinite. 
Here's what I have as a proof, but I'm not really sure if it's enough. 
Let $X \subseteq Y$ and $X\neq Y$, this must mean that $|X| < |Y|$ which also implies there's an injection. 
there's a bijection from $Y$ to $X$ implies |X| = |Y|.
Suppose that Y is infinite. Then it can either be countable infinite, or uncountable infinite. Let Y be countable infinite. 
Y = {$y_1$, $y_2$,...,$y_n$} and let X = {$x_1$,$x_2$,...,$x_m$}
if $ m \neq n$ then there cannot be a surjection, which contradicts the fact that f defines a bijection. Now suppose $m = n$ and Y is countable infinite. This fact holds as long as X is countable infinite. 
Now suppose that Y is uncountable infinite. Then there can be a bijection as long as X is uncountable infinite. 
Another strategy? 
Could this proof be as simple as showing a contradiction? 
Assumptions I'll use: 
$f: [m] \rightarrow [n]$ defines a bijection then $m=n$
Suppose that Y is finite. if $f: Y \rightarrow X$ is a bijection, we know that |X| = |Y|, but |X| < |Y|, which implies that $m \neq n$ which is a contradiction. 

Using fundamental theorems of cardinality, if $Y$ happens to be finite then so is $X$. Then, from the inclusion and using the fact that $Z_1\subset Z_2$, $|Z_1|=|Z_2|$ implies $Z_1=Z_2$ if $Z_1$ and $Z_2$ are finite, we see that $|X|<|Y|$ on the other hand, using the bijection $f$ we get $|X|=|Y|$ which is a contradiction. 
I would like to highlight another proof which is better, IMHO, because we are not reasoning ad absurdum. 
Clearly $f(Y)\subseteq Y$ and $f(Y)\neq Y$. Now both affirmation are necessarily stable by applying $f$ (since $f$ is a bijection) so :
$$f^2(Y)\subseteq f(Y)\text{ and } f^2(Y)\neq f(Y) $$
and so on... so that for all $n\geq 0$ :
$$f^{n+1}(Y)\subseteq f^n(Y)\text{ and } f^{n+1}(Y)\neq f^n(Y) $$
This means that you can find $x_n\in f^{n+1}(Y)-f^n(Y)$ for each $n$. Clearly if $n<m$ then $x_m\notin f^{n+1}(X)$ so $x_n\neq x_m$. Finally we constructed an infinite sequence $(x_n)$ where each pair of elements is distinct.
Of course I use here the countable axiom of choice, if you don't want to use this, just construct $(x_1,...,x_N)$ for each $N>0$ (you just make a finite number of choice) and conclude that $N\leq |Y|$ for each $N>0$. 
