This is from Conway's on numbers and games:
$x+(-x)=0$. We have to show $x+(-x)\geq 0$ and $x+(-x )\leq 0$. If say $(x+(-x))\ngeq 0$, we should have some $(x+(-x))^R\leq 0$, that is $x^R+(-x)\leq 0$ or $x+(-x)^L\leq 0$. But these are false, since we have by induction $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$.
I know that $x^R+(-x)\leq 0$ is a right option of $(x+(-x))$ so if its negative the number is. I don't understand where the "or $x+(-x)^L\leq 0$" come from. And why do we have $x^R+(-x)^R\geq 0$, $x^L+(-x)^L\geq 0$ by induction?
Where does $x+(-x)^L\le0$ come from?
It is a typo that your question has that my copy of the book does not, possibly related to confusion about notational conventions.
My copy says $x+-x^L$, not $x+(-x)^L$. This is because the right options of $x+(-x)$, that is, the ones denoted by "$(x+\,-x)^R$" in the book, have either the form $x^R+(-x)$ or the form $x+(-x)^R$. And by the definition of $-x$, something like $(-x)^R$ is like $-(x^L)$. By the conventions related to order of operations in regular arithmetic (since exponentiation has higher precedence than multiplication by $-1$), we write $-x^L$ to mean $-(x^L)$ and not $(-x)^L$.
Why do we have things by induction?
This is related to the paragraph just before PROPERTIES OF ORDER AND EQUALITY.
In this case, the implied inductive assumption is that $y+\,-y\ge0$ is true for all of the options of $x$, so that $x^R+\,-x^R\ge0$ and $x^L+\,-x^L\ge0$.
Where's the rest of the proof?
In case it wasn't clear, when the book says "say, $x+\,-x\ngeq0$", it's implying that we're only proving $x+\,-x\ge0$ and that the proof of $\le0$ is analogous.