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?

Yes, $ $ your proof is correct. $ $ You have $\ a = \color{#c00}{(x\!-\!1)/2}\in\Bbb Z\ $ so eliminating $\,a\,$ the proof becomes
$$\begin{align} &\dfrac{x^2}2 - \dfrac{x}2\, =\, \dfrac{x(\color{#c00}{x\!-\!1})}{\color{#c00}2}\in \Bbb Z\\[.2em]
\Rightarrow\ \ \ &\dfrac{x^2}2\in\Bbb Z\iff \dfrac{x}2\in\Bbb Z\\[.2em]
\Rightarrow\ \ \ & \ \ \ \ 2\nmid {x^2}\iff\ \  2\nmid x\\[.2em]
{\rm i.e.}\  \ \ & \ x^2\  {\rm odd}\ \iff x\ \  {\rm odd}
\end{align}\qquad\qquad\qquad\qquad$$
