Every affine algebraic set has finitely many irreducible components

The ideal $I(X)$ is radical by [L1], so [L2] yields minimal prime ideals ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ over $I(X)$ with $I(X)={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_m\mathrm{.}$ Applying [L1] again, $X=V(I(X))=V({\mathfrak{p}}_1)\cup \cdots \cup V({\mathfrak{p}}_m),$ and each $V({\mathfrak{p}}_i)$ is irreducible.

This union is irredundant. Indeed, if $V({\mathfrak{p}}_i)\subseteq V({\mathfrak{p}}_j)$, then [L1] reverses inclusions and gives ${\mathfrak{p}}_j\subseteq {\mathfrak{p}}_i$. Minimality of the ${\mathfrak{p}}_i$ over $I(X)$ forces equality, so distinct terms cannot contain one another.

Let $Y\subseteq X$ be any irreducible closed subset. Then $Y$ is an affine algebraic set, so $Y=V(\mathfrak{q})$ for some prime ideal $\mathfrak{q}$ containing $I(X)$ by [L1]. Since the ${\mathfrak{p}}_i$ are precisely the minimal primes over $I(X)$, some ${\mathfrak{p}}_i$ is contained in $\mathfrak{q}$, hence $Y=V(\mathfrak{q})\subseteq V({\mathfrak{p}}_i)$. Therefore every irreducible closed subset of $X$ lies in one of the $V({\mathfrak{p}}_i)$.

Step 2.2 shows that the maximal irreducible closed subsets of $X$ are exactly the sets $V({\mathfrak{p}}_i)$. Hence the family from step 1.1 is uniquely determined up to order and is precisely the irreducible-component decomposition of $X$.