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

Let $x = 2k + 1 \in \mathbb{Z}$ be odd. So, $$ x^{2} = (2k + 1)^2 = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$$ Let $t = 2k^{2} + 2k \in \mathbb{Z}$, thus: $ x^{2} = 2t + 1$.

Your written proof is correct. FYI, here is another proof technique which, although it's far more than you need in your case as Bill's comment indicates, is somewhat shorter and, perhaps, of some use to you, such as for other more complicated related problems.

Since $x$ is odd, this means it has no factors of $2$. By the Fundamental theorem of arithmetic, $x^2$ has the same prime factors as $x$, just twice as many of each of them and, thus, also no factor of $2$. As such, it is also odd.