Simple question about square of a supremum

In case you want to show this rigorously, you will need to show that $\sup \{a^2\} \leq (\sup \{a\})^2$ and vice-versa.

To show one direction, note that $\sup \{a\} \geq a$ for all $a$, hence $(\sup \{a\})^2 \geq a^2$ for all $a$, so $(\sup \{a\})^2$ is an upper bound for $\{a^2\}$, hence it is greater than the least upper bound, i.e. $(\sup\{a\})^2 \geq \sup\{a^2\}$.

To show the other direction, let $N$ be large enough such that $\sup \{a\} - \frac 1N > 0$ (if $\sup \{a\} = 0$ and all $a$ are non-negative, then it is clear what the set is and the result itself). Now for any $n>N$, the quantity $\sup \{a\} - \frac 1n$ is not an upper bound of $\{a\}$, Hence, there is some $a_n$ such that $a_n > \sup\{a\} - \frac 1n$. Square both sides(note that by non-negativity of both sides, this preserves sign) to see that $a_n^2 > \frac 1{n^2} + (\sup \{a\})^2 - \frac{2\sup \{a\}}{n}$. Now, by definition of supremum, we have $$\sup \{a^2\} \geq a_n^2 > \frac 1{n^2} + (\sup \{a\})^2 - \frac{2\sup \{a\}}{n}$$

This applies for all $n>N$. Since $(\sup \{a\})^2$ is a bounded quantity, letting $n \to \infty$, we see that $\sup \{a^2\} \geq (\sup \{a\})^2$. Hence, equality follows.

Note the trick played in the second half of the proof. You will need to use it repeatedly and become comfortable with it.