If $B\subset A$ and $f:A\to B$ is injective prove it's a bijection between $A$ and $B$ I want to show that if $B\subset A$ and $f:A\to B$ is an injective function then there's a bijection between $A$ and $B$.

I believe my "proof" is wrong, I probably use too much "intuition" when I try to solve it. But hopefully I will get a better feeling if someone tells me where/what I do wrong and help me. :) That said, a friend to me "solved" another problem 
"if $f:A\to C$ is an injective function, and $g:C\to A$ is an injective function, then there is a bijection between $A$ and $C$". 
She argued like this " for all $a$ in $A$ we can find an element $f(a)$ in $C$ and for all $c$ in $C$ we have an element $g(c)$ in $A$. If f(a)=c then we must have g(c)=a. This holds for all $a$ and $c$. So each $a$ maps to exactly one $c$ and each $c$ maps to exactly one $a$". 
That is the Schröder–Bernstein theorem though, I have seen the proof, so I could directly tell that it's a wrong proof. I would probably not argue in the exactly same way but probably in a similar fashion. On the other hand I cannot really tell why this doesn't prove the fact either. Oh well, here comes my proof, it's a similar argument so I guess I'm wrong :)
Proof:
Since I already know it's injective, I just have to show it's surjective. We have that $B\subset A$, that is, every element of $B$ is in $A$. Because of this we can for every element $b\in B$ find an element $a\in A$ such that $b=f(a)$. That is, $\forall b\in B\exists a\in A:\textbf{ }b=f(a)$. But that is the definition of surjection. Hence, there exists a bijection between $A$ and $B$ since f is injective.
I bet I've forgot to mention something now, which I found important to mention, but unfortunately I have forgotten it. Hopefully I will remind myself. Thanks for your help :)
 A: It doesn't hold that $f$ gives the bijection. Consider $A=(0,2),B=(0,1)$ and $f(x)=\dfrac{x}{3}.$
The problem in your proof: Each element of $B$ is an element of $A,$ but this doesn't mean that an element of $B$ has a preimage. Think of the above example.
A: Just argue like this:
$B\subset A$ gives an injection $i:B \rightarrow A$ (the standard inclusion: every $b \in B$ is sent to itself). By hp, you are given $f: A \rightarrow B$ injective, so that by Cantor-Berstein you know $A$ and $B$ are in bijection.
A: Well, your friend is wrong.  $f(a)$ maps to exactly one $c$ and $g(c)$ maps to exactly one $b \in A$ but there is no reason in any of the pluperfect hells to assume that $b$ is the same as the original $a$.
Anyway you comment isn't true.  Let $f: (0,1) \rightarrow (0,1/2)$ $f(x) = \frac 1 4 x$.  There is no $x \in (0,1)$ such that $f(x) = 1/3$.
Of course if the statement was $f: A \rightarrow f(A) \subset A$.... that'd be a different story.
A: Just because $A \subset B$ doesn't necessarily mean that $\forall b \in B \exists a \in A \colon b = f(a)$; for instance: $f \colon \mathbb N \rightarrow \mathbb N*, f(x) = 2 \times x + 1$. $ \mathbb N* \subset \mathbb N$. The function is injective, but there exists $b$ such that $ \nexists a \in A \colon b = f(a)$, for instance $b = 0$. That's the problem with the proof(essentially, you implicitly assumed that the function is surjective). A simpler proof would be: let $g \colon B \rightarrow A$ so that $g(x) = x$. The function is obviously injective. Therefore we have both $f \colon A \rightarrow B$ and $g \colon B \rightarrow A$, both injective. Therefore, by the Schröder–Bernstein theorem, there must exist a bijection between $A$ and $B$.
