Proof: If $x$ is odd, then $x+2$ is odd. I'm fairly new to writing proofs so any advice can help. 
I'm asked to prove the following statement: "If $x$ is odd, then $x+2$ is odd". Here is my proof: 
We will prove this by contraposition: if $x+2$ is not odd, then $x$ is not odd. 
Let there be an integer $k$ such that $x+2 = 2k$.
Thus,
\begin{align}
x & = 2k-2 \\
 & = 2(k-1)
\end{align}
Then $x = 2(k-1)$ is an even number. 
Since the contrapositive is true, the statement "If $x$ is odd, then $x+2$ is odd" is true by logical equivalency. 

The problem is: I don't know if my proof is enough or how to properly tackle them. Any advice? 

An integer $m$ is even if and only $2 \mid m$; otherwise, $m$ is odd.
If $x + 2$ is even, then
$2 \mid x + 2; \tag 1$
clearly,
$2 \mid 2; \tag 2$
thus
$2 \mid (x + 2) - 2 = x; \tag 3$
that is, $x$ is even, contrary to hypothesis; thus, $x + 2$ cannot be even, hence it is odd.
By the way, our OP's proof, being essentially the same as mine, is fine.
