Spectrum is compact

Statement
The spectrum of a commutative unital ring is always a compact space.

Proof
We shall prove the statement using the finite intersection property formulation of compactness.

Given: $$R$$ is a commutative unital ring, and $$X = Spec(R)$$ is its spectrum. Suppose $$V_i$$ is a collection of closed subsets (where $$i \in I$$, an indexing set) such that any finite intersection of the $$V_i$$s is nonempty.

To prove: The intersection of all the $$V_i$$s is nonempty.

Proof: Let $$\mathcal{I}(V)$$ denote the radical ideal corresponding to the closed set $$V$$ under the Galois correspondence between a ring and its spectrum.

Clearly, we have, by definition:

$$\mathcal{I}(\bigcap_i V_i) = \bigoplus \mathcal{I}(V_i)$$

(the right side denotes the ideal generated in $$R$$, and not the abstract direct sum).

In other words, a prime ideal contains all the $$\mathcal{I}(V_i)$$s if and only if it contians their sum.

Suppose the intersection of the $$V_i$$s was empty. Then the left side would be the whole ring $$R$$, hence so would the right side. In other words, we'd have that $$R$$ is the sum of the $$\mathcal{I}(V_i)$$s.

But this would imply that $$1 \in R$$ is in the sum of finitely many $$\mathcal{I}(V_i)$$s, and that in turn would yield that the intersection of those finitely many $$V_i$$s must be empty, contradicting the assumption of finite intersection property.