Irreducible closed subsets of a scheme corresponds to points

I have posted an answer here for the case of an affine scheme, but I got stuck when I tried to generalize the argument to schemes.

My thoughts

Consider a point $p$ in the scheme, its closure in the scheme is an irreducible closed subset. This gives us one direction of the correspondence.

Let $C$ be an irreducible closed subset of the scheme, pick an affine neighborhood $U$ that intersects nontrivially with $C$. Then the intersection is a closed subset of $U$ which decomposes into finite union of irreducible closed subsets of $U$ by Noetherian property of $U$. This is where I got stuck, and don't know how to proceed from here.

Hint: First the closure of a point is always irreducible, there's no need to pass through an affine neighborhood for that.

For the other direction if $C$ is an irreducible closed set and $U$ is open then show that $C \cap U$ is an irreducible closed subset of $U$.

Now $U \cap Z$ has a unique generic point when $U$ is affine. Show that if $V$ is also affine and $V \cap Z$ is nonempty then the generic point of $U \cap Z$ lies in $V$ and hence equals the unique generic point of $V \cap Z$.

Finally use the fact that affine opens form a base for the topology to show that the common generic point for all nonempty $U \cap Z$ is a generic point for $Z$, necessarily a unique one.

I am going to write my answer to this just in order to collect everything in one place.

Definition. Let $ X $ be a topological space. A point $ \eta \in X $ is said to be a generic point if $ \eta \in \overline { \left \{ x \right \} } $ implies $ \eta = x $.

We easily see that the property of being generic is local on $ X $ i.e. if $ \eta \in U \subset X $ where $ U $ is open, then $ \eta $ is generic in $ X $ if and only if $ \eta $ is generic in $ U $.

Lemma 1. If $ X $ is an integral affine scheme, then $ X $ has a unique generic point $ \eta $. It is characterized by the property that its closure is all of $ X $ or equivalently, it is contained in every open subset of $ X $.

Proof: Let $ X = $ Spec $ A $. Then, $ \eta = 0 $ is a prime ideal as $ A $ is integral. If $ \mathfrak{p} \in $ Spec $ A $, such that $ \eta \in \overline { \left \{ \mathfrak{p} \right \} } = V( \mathfrak{p} ) $, then $ \eta \supset \mathfrak{p} $, whence $ \eta = \mathfrak{p} = 0 $. Thus, $ \eta $ is a generic point. Notice that $ \overline { \left \{ \eta \right \} } = V( 0 ) = X $. If $ \eta' $ is also a generic point, then as $ \eta' \in \overline { \left \{ \eta \right \} } $, we have $ \eta ' = \eta $, whence $ \eta $ is unique. The characterization property is obvious. $ \square $

Lemma 2. If $ X $ is an integral scheme, then $ X $ has a unique generic point $ \eta $. It is characterized by the property that $ \overline { \left \{ \eta \right \} } = X $ or equivalently, it is contained in every open subset of $ X $.

Proof: Saying that $ X $ is integral is the same as saying that $ X $ is reduced and irreducible. Let $ U = $ Spec $ A $ be an open affine in $ X $. Since $ X $ is irreducible, $ U $ is dense in $ X $. As $ A $ is integral, $ \eta = 0 \in $ Spec $ A $ is a generic point of $ U $ by Lemma 2, and hence a generic point of $ X $. This proves the existence. If $ \eta ' \in X $ is also generic, then if $ U ' \subset X $ is an open affine around $ \eta ' $, and $ V \subset U \cap U ' $ is an open affine, then $ \eta , \eta' \in V $ by the characterization property in Lemma 2. Then, both $ \eta $ and $ \eta ' $ are generic points of $ V $, and thus $ \eta = \eta ' $. This proves uniqueness. The characterization property is clear by construction of $ \eta $. $ \square $

Lemma 3. If $ X $ is a topological space, $ C $ is an irreducible subset of $ X $, then the closure $ \overline{C}$ of $ C $ in $ X $ is irreducible. In particular, the closure $ \overline { \left \{ x \right \} } $ of any point $ x \in X $ is irreducible.

Proof: If $ \overline{C} $ is not irreducible, then there are two non-empty open subsets $ U $ and $ V $ of $ \overline{C} $ such that $ U \cap V = \emptyset $. Now, $ U \cap C$ and $ V \cap C $ are open and disjoint subsets of $ C $. If either is empty, say $ U \cap C $, then the points in $ U \subset \overline{C} $ cannot be in the closure of $ C $ in $ \overline{C} $ which is $ \overline{C} $. $ \quad \quad \quad \quad $ $ \square $

Now, suppose that $ X $ is a scheme. By Lemma 3 above, each point gives an irreducible closed subset of $ X $. Conversely, if $ C $ be an irreducible closed subset, then, as $ C $ is closed, it has a unique reduced induced sub-scheme structure, and as $ C $ is irreducible, this scheme structure is integral. Thus, by Lemma 2, it has a unique generic point.