**Proof.**
By Cohomology of Spaces, Lemma 67.20.3 we see that $A = H^0(X, \mathcal{O}_ X)$ is a finite dimensional $k$-algebra. This proves (1).

Then $A$ is a product of local rings by Algebra, Lemma 10.52.2 and Algebra, Proposition 10.59.6. If $X = Y \amalg Z$ with $Y$ and $Z$ open subspaces of $X$, then we obtain an idempotent $e \in A$ by taking the section of $\mathcal{O}_ X$ which is $1$ on $Y$ and $0$ on $Z$. Conversely, if $e \in A$ is an idempotent, then we get a corresponding decomposition of $|X|$. Finally, as $|X|$ is a Noetherian topological space (by Morphisms of Spaces, Lemma 65.28.6 and Properties of Spaces, Lemma 64.24.2) its connected components are open. Hence the connected components of $|X|$ correspond $1$-to-$1$ with primitive idempotents of $A$. This proves (2).

If $X$ is reduced, then $A$ is reduced (Properties of Spaces, Lemma 64.21.4). Hence the local rings $A_ i = k_ i$ are reduced and therefore fields (for example by Algebra, Lemma 10.24.1). This proves (3).

If $X$ is geometrically reduced, then same thing is true for $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ (see Cohomology of Spaces, Lemma 67.11.2 for equality). This implies that $k_ i \otimes _ k \overline{k}$ is a product of fields and hence $k_ i/k$ is separable for example by Algebra, Lemmas 10.43.1 and 10.43.3. This proves (4).

If $X$ is geometrically connected, then $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ is a zero dimensional local ring by part (2) and hence its spectrum has one point, in particular it is irreducible. Thus $A$ is geometrically irreducible. This proves (5). Of course (5) implies (6).

If $X$ is geometrically reduced and connected, then $A = k_1$ is a field and the extension $k_1/k$ is finite separable and geometrically irreducible. However, then $k_1 \otimes _ k \overline{k}$ is a product of $[k_1 : k]$ copies of $\overline{k}$ and we conclude that $k_1 = k$. This proves (7). Of course (7) implies (8).
$\square$

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