To proof I said, Let $A$ be an ordered integral domain. Let $x\in A$ such that $x^2+e=0$. Then $x^2=-e$. Then $x\neq0$ would imply that $x^2\in A^+$ where $A^+$ is all positive elements of A and would imply that $-e,e\in A^+$. Thus, $-e+e=0$ would be an element of $A^+$ and would be a contradiction because $0\notin A^+$. Therefore, $x^2+e=0$ has no solution in an ordered integral domain.
I was wondering if this is a okay proof or what I could improve on. Thank you.
You proof is good but you are making a few assumptions that must be proven.
You are assuming that if $a \ne 0$ then $a^2 > 0$. That is true but in is not axiomatic and must be proven.
You are assuming that $e \in A^+$. That is true but again is not axiomatic and must be proven.
And you are assuming $-e,e \in \mathbb A^+$ is a contradiction? Why is that? You seem to be applying that if $a > 0, b > 0$ then $a+b > 0$? Is that true? (Yes, it is... although maybe I'm not sure what definition of $a > 0$ is that you are working with... that could be on me.)
But yes... your proof is correct and if we can show $a^2 > 0$ when $a \ne 0$ and that $e\in A^{-a}$ and $a, -a \in A^+$ is always a contradiction then your proof is complete.