prove that $\alpha\le \beta \iff \exists !\gamma(\alpha+\gamma=\beta)$ where $\alpha,\gamma,\beta$ are ordinals. My first attempt at a proof is as follows
$(\rightarrow)$ suppose that $\alpha\le \beta$ then we have two cases one when $\alpha<\beta$ and one when $\alpha=\beta$ so when $\alpha=\beta$ clearly $\alpha+0=\beta$ and when $\alpha<\beta$ we get that $\alpha+\delta=\beta$ for some $\delta < \beta$ hence if $\alpha\le \beta \implies \exists ! \gamma(\alpha+\gamma=\beta)$
$(\leftarrow)$ suppose $ \exists !\gamma(\alpha+\gamma=\beta)$ then there are again only two cases when this holds, one being $\alpha=\beta$ and one when $\alpha<\beta$ so we must have that $\alpha\leq\beta$ for this condition to be true.
My questions are; Is my proof correct?, If not how would i correct it, and shoult it be that $\delta \le \beta$ not $\delta < \beta$ because $\alpha$ could equal $0$?. [or do i need to use transfinite induction on $\gamma$?].
If you've already proved that ordinal addition is increasing in the right argument (that is, that $\alpha+\beta<\alpha+\gamma$ whenever $\beta<\gamma$), then your second half is just fine. However, you haven't actually proved what you're supposed to prove for the non-equal case of the first half.
It seems (though I could be wrong) that you've already proved that if $\alpha<\beta,$ then there is some $\delta>0$ such that $\alpha+\delta=\beta.$ If so, then what you must prove, here, is uniqueness. This immediately follows from ordinal addition being increasing in the right argument, though. If you haven't proved such a result, then you still need to, at which point uniqueness follows readily from the other result I mentioned.
You are quite correct about one issue: it should be $\delta\le \beta,$ and not only because we could have $\alpha=0.$ For example, taking any $\alpha<\omega,$ we have $\alpha+\omega=\omega.$ There is no need for transfinite induction in this proof, but (if memory serves) one does need it to prove the two underpinning results I mentioned.