Prove that if $x$ is odd, then $x^2$ is odd 
Prove that if $x$ is odd, then $x^2$ is odd

Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\frac{x}{2} = x(a+0.5) = xa + \frac{x}{2}$$
$xa \in \mathbb Z$ and $\frac{x}{2} \notin \mathbb Z$, hence $xa + \frac{x}{2}$ is not a integer. And since $xa + \frac{x}{2} = \frac{x^2}{2}$, it follows that $x^2$ is not divisible by two, and thus $x^2$ is odd.
Is it correct?

Your proof is correct.  
You could also prove by contrapositive:  
if $x^2$ is even, then $2|x^2=x\cdot x,$ so by Euclid's lemma $2|x,$ so $x$ is even.  
If you don't like that, you could say $x$ is odd, and the product of two odd numbers is odd, 
so $x^2=x\cdot x$ is odd.
If you don't like that, you could argue as follows:
if $x$ is odd, then $x-1=2k$ and $x+1=2k+2$ with $k \in\Bbb Z$, 
so $x^2=(x-1)(x+1)+1=(2k)(2k+2)+1=2(k(2k+2))+1$ is odd.
