$|\mathbb{Z}|=|\mathbb{N}|$ 
Hello all, if you could give some critique on my proof I would be grateful.

Show that $|\mathbb{Z}|=|\mathbb{N}|$ for,
$$\displaystyle f(n) = \begin{cases} 2|n|+1, \text{if} & n\leq0 \\ 2n, \text{if} & n>0\end{cases}$$
By the definition of cardinality (def. 2.10b) we note that if a function is bijective, cardinality of the sets equal to one another. So, we want to show that $f(n)$ is bijective. To prove that $f:\mathbb{Z}\rightarrow \mathbb{N}$ injective, we must show $\forall n_1,n_2\in \mathbb{Z}$, if $f(n_1)=f(n_2)$ then $n_1=n_2$ for both conditions of $f(n)$, so,
$f(n)=2|n|+1 $, if $ n\leq 0$
$2|n_1|+1 = 2|n_2|+1$
$2|n_1|=2|n_2|$
$|n_1|=|n_2|$
and
$f(n)=2n$, if $n>0$
$2n_1=2n_2$
$n_1=n_2$
Thus we see that $f(n)$ is clearly injective. To prove $f:\mathbb{Z}\rightarrow \mathbb{N}$ is surjective, meaning $\exists n\in \mathbb{Z},\forall r\in \mathbb{N}$ s.t. $f(n)=r$, again for both conditions,
$f(n)=2|n|+1 $, if $ n\leq 0$ 
$r=2|n|+1$
$r-1=2|n|$ 
$|n|=\frac{r-1}{2}$
and 
$f(n)=2n$, if $n>0$
$r=2n$
$n=\frac{r}{2}$
Thus we see that for both conditions $f(n)=r$, and function $f:\mathbb{Z}\rightarrow \mathbb{N}$ is surjective. Hence, we prove $f:\mathbb{Z}\rightarrow \mathbb{N}$ is bijective as it is injective and surjective. Therefore, by definition of cardinality, we say that cardinality of $\mathbb{N}$ is equal to $\mathbb{Z}$. $\blacksquare$

You are very very close.  
But:
You say "Now, suppose f(n) is injective, meaning ∀n1,n2∈Z, if f(n1)=f(n2) then n1=n2 for both conditions of f(n), so,..."
This is what you want to prove. You are not "supposing" it.  ANd the "so...." does not follow as a result.  What follows is your attempt to prove it.
ANd then 
$f(n) = 2|n| + 1$ if $n \le 0$
$2|n_1| + 1 = 2|n_2| + 1$.  But you do not know if $n_1, n_2$ are positive or negative.  You must conisder all options that $f(n_1) = f(n_2)$ even if they are different signs.
And then you get:
$|n_1| = |n_2|$ therefore
$n_1 = n_2$.  That is not true.  It is possible that $n_1 = -n_2$.
You might need to do cases:
$f(n_1) = f(n_2)$
Case 1: $n_1 \le0; n_2\le 0$.
$2|n_1|+1 = 2|n_2| + 1$
$|n_1| = |n_2|$ but as $n_1 \le 0$ and $|n_2 \le 0$ 
$-n_1 = - n_2$ and $n_1 = n_2$.
Case 2: $n_1 > 0; n_2>0$.
Then $2|n_1| = 2|n_2|$
$|n_1| = |n_2|$ but as both are positive
$n_1 = n_2$.
Case 3: $n_1 \le 0$ and $n_2 > 0$
Then $2|n_1| + 1= 2|n_2|$.
This is an odd number equaling an even number.  This is impossible.
Case 4: $n_1 > 0; n_2 \le 0$
This is no different essentially than case 3.
...
Notice it might be easier if you had done
$f(n) = -2n + 1$ if $ n \le 0$ and $f(n) = 2n$ if $n > 0$.
Then we could say $f(n_1) = f(n_2) $ means either
1) $-2n_1 + 1 = -2n_2 + 1$
2) $2n_1 = 2n_2$
3) $-2n_1 + 1 = 2n_2$
4) $2n_1 = -2n_2 + 1$.
1) and 2) both reveal $n_1 = n_2$.
3) $n_2 = -n_1 + \frac 12$.  Which is impossible as the RHS is not an integer.
4) $n_1 = -n_2 + \frac 12$.  Ditto.
.... oh.... Your surjectivity is kind of a mess.
To prove surjectivity you choose and arbitrary $r \in \mathbb N$ and prove there must be an $n \in \mathbb Z$ so that $f(n) = r$.  You have to start with the $r$ and prove there is an $n$ that goes to it.  You can't start with the $n$.  You get $n = |\frac {r-1}2|$.  Well, so?  How do you know that that is even an integer.  It isn't if $r$ is even.  How do you know $r$ is odd?
So suppose $r \in \mathbb N$.  Either $r$ is even.  Or $r$ is odd$.
If $r$ is even let $n = \frac r2$.  And $n > 0$ we know $n =\frac r2 > 0$.
So $f(n) = 2|n| = 2|\frac r2| = |r| = r$.
If $r$ is odd, let $n = -\frac {r-1} 2$.  So $r \ge 1$ then $n = -\frac {r-1}2 \le 0$.
So $f(n) = 2|n| + 1= 2|-\frac {r-1}2| + 1 = 2*\frac {r-1}2 + 1 = (r-1) + 1 = r$.
So for any $r \in \mathbb N$ there will be an $n\in Z$ so that $f(n) = r$.
Again as youare defining $f$ in terms of whether $n$ is positive or non-positive, you don't/shouldn't define $f$ in terms of absolute values.
If $f(n) = 2|n| + 1$ then we know $n \le 0$ so just say $f(n) = -2n + 1$.
And if $f(n) = 2|n|$ when we know $n > 0$ so just say $f(n) = 2n$.
Also as Yves Daoust suggest:
consider defining $g: \mathbb N \to \mathbb Z$ as $g(n) = \frac n2$ if $n$ is even, and $g(n) = -\frac {n-1}2$ if $n$ is odd.
You'll find this direction is a bit easier and more natural (although your direction is also correct.)
If $n$ if even then $g(n) = \frac n2 > 0$.  If $n$ is odd then $g(n) = -\frac {n-1} 2 \le 0$.
So if $g(n) = g(m) > 0$ then $m,n$ are both even and $\frac m2 = \frac n2$ so $m = n$.
And if $g(n) = g(m) \le 0$ then $m,n$ are both odd and $-\frac {m-1}2=-\frac {n-1}2$ so $m = n$.
So $g$ is injective.
And for any $r \in \mathbb Z; r >0$ then $n = 2r \in \mathbb N$ and $g(n) = \frac {2r}2 = r$.
And for any $r \in \mathbb Z; r \le 0$ then $n = -2r + 1 \in \mathbb N$ and $g(n)=- \frac {(-2r + 1)-1}2= r$.
So $g$ is surjective.
