Let $C$ be curve, so a $1$-dimensional, separated $k$-scheme of finite type).
My question is how to see that the underlying topological space of $C$ is a Noetherian space, so satisfies the descending chain condition for closed subsets.
My attempts:
Obviously $C$ is locally noetherian:
Because $C$ is $k$-scheme of finite type so we can find for each $c∈C$ a wlog affine open neighborhood $U_c=Spec(R)$ of $c$ such that $R=k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $⇔ Spec(R)$ noetherian works only for affine schemes.
Generally $C$ is not affine so I hornestly don't know how to show that $C$ is a noetherian space in satisfying way. One way to see it is would be to embedd it in a $\mathbb{P}^n$. But I find it a bit overkill like so would like to prefer a more elementary argument basing on the definition of a curve as given above.
Since $C$ is finite type over $k$, it is quasicompact, so it is covered by finitely many affine open sets (each of which are Noetherian). It follows immediately that $C$ is Noetherian (given any descending sequence of closed sets, intersect it with each affine open set and it must eventually stabilize on each one).