Problem: Given a monomial ordering in polynomial ring of n variables. Prove that every descending sequence of monomials terminates.
My attempt: Let $>$ be a monomial ordering on $\mathcal{M}$, the set of all monomials in $k[x_1,\dots,x_n]$. Hence $>$ be a well-ordering relation on $\mathcal{M}$, i.e. every nonempty subset of $\mathcal{M}$ has minimal element. Suppose there is an infinite descending $x^{\alpha(1)} > x^{\alpha(2)} > x^{\alpha(3)} > \cdots$ of monomials. There is some $\alpha(n)$ such that $x^{\alpha(n)} = x^{\alpha(n+1)} = x^{\alpha(n+2)} = \cdots$. Q.E.D
Your proof is OK, but you can add some details: the fact that $>$ (or normally we would say $<$) well-orders $\mathcal{M}$ is due to the fact that we have only finitely many variables $x_n$. In the concluding line, there is some $n$ such that $x^{\alpha(n)}=\min_<\{x^{\alpha(i)}:i\in\mathbb{N}\}$. From this it follows like you said that $x^{\alpha(n)}=x^{\alpha(n+1)}=\dots$ contradicting your assumption. Note that an equivalent definition of a well-ordering relation is that it does not contain infinite descending sequences.