Use a proof by contradiction. Assume that $\displaystyle s^2 > 3$ and show that there exists another rational $\displaystyle q < s$ with $\displaystyle q^2 > 3$. This contradicts the assertion that $\displaystyle s$ is the infimum of the set.
Spoiler:
Once you have constructed such a $\displaystyle q$, you will probably be able to use the same technique to prove that there is no rational that is the supremum of the complement of the given set, thus proving that $\displaystyle \sqrt{3}$ is irrational.
The problem's claim is false, because $\displaystyle \inf\{ x \in \mathbb{Q} \ | \ x^2 > 3 \} = -\infty$.
For example, $\displaystyle -20 \in \mathbb{Q}$, and $\displaystyle (-20)^2 = 400 > 3$, so $\displaystyle -20 \in \{ x \in \mathbb{Q} \ | \ x^2 > 3 \}$,
and thus $\displaystyle s = \inf \{ x \in \mathbb{Q} \ | \ x^2 > 3 \} \le -20$, and so $\displaystyle s^2 \ge 400$. So it's just false that $\displaystyle s^2 \le 3$.
Of course, since negative infinity isn't a number, even what I wrote isn't correct - but you see the problem. (If you define the arithmetic of the extended reals (extended to plus and minus infinity) in the natural way, then what I wrote is ok.)
I suspect what was intended was one of these:
Let $\displaystyle s = \inf\{ x \in \mathbb{Q} \ | \ x \ge 0 \text{ and } x^2 > 3 \}$. Prove $\displaystyle s^2 \le 3$. (It will be that $\displaystyle s = \sqrt{3}$.)
or (basically equivalently)
Let $\displaystyle s = \inf\{ |x| \ | \ x \in \mathbb{Q} \text{ and } x^2 > 3 \}$. Prove $\displaystyle s^2 \le 3$. (It will be that $\displaystyle s = \sqrt{3}$.)
or (this is the easiest to prove)
Let $\displaystyle s = \inf\{ |x| \ | \ x \in \mathbb{Q} \text{ and } x^2 > 3 \}$. Prove $\displaystyle s^2 \ge 3$. (It will be that $\displaystyle s = \sqrt{3}$.)
or (less likely)
Let $\displaystyle s = \inf\{x \in \mathbb{Q} \ | \ x^2 < 3 \}$. Prove $\displaystyle s^2 \le 3$. (It will be that $\displaystyle s = -\sqrt{3}$.)
or (much less likely, since here s won't be real number, and this isn't finding the square root of 3 via real number completeness)
Let $\displaystyle s = \inf\{x \in \mathbb{Q} \ | \ x^2 > 3 \}$. Prove $\displaystyle s^2 \ge 3$. (It will be that $\displaystyle s = -\infty$.)