The answer was given $-\sqrt{2}$, but $A$ contains only rationals, so $\sqrt{2}\notin A$ and $-\sqrt{2}\notin A$ as well. Thus the set $A$ shrinks to the set of all rational numbers, between $-1$ and $\sqrt{2}$.
My answer, when I solved the problem, was $\inf A= -1$. The proof goes like
Since $-1< x\,\,\forall x\in A$, therefore $-1$ is an lower bound of $A$. Let's choose any arbitrary $\varepsilon>0$. Since $\varepsilon>0$, we have from the Archimedean property that $\exists \,p\in\mathbb{N}$ such that $\dfrac{1}{p}< \varepsilon\Rightarrow -1+\dfrac{1}{p}<-1+\varepsilon$.
Since $\dfrac{1}{p}\in \mathbb{Q}\Rightarrow -1+\dfrac{1}{p}\in\mathbb{Q}$ and $-1+\dfrac{1}{p} > -1$, thus $-1+\dfrac{1}{p}\in A$ and $-1< -1+\dfrac{1}{p}< -1+\varepsilon\Rightarrow \inf A= -1$
It is not possible for $x^2 = 2$ if $x \in \mathbb{Q}$, so the set in question is $\{x \in \mathbb{Q} : -1 < x < \sqrt{2}\}$. Clearly, the infimum is $-1$ since $-1$ is a lower bound and if we had some lower bound $M > -1$, we would have some rational $x$ such that $-1 < x < M$ by the density of the rationals.
In other words, your answer is correct.