A regular point lies on one irreducible component

Fix $x$. By the AC assumption [F1] and [F2], choose an affine open neighbourhood $U=\mathrm{Spec}A$ of $x$ with $A$ Noetherian. Write $x$ as the prime $\mathfrak{p}\subset A$. The open-scheme structure in [F3] and the affine stalk calculation [F4] identify ${\mathcal{O}}_{X,x}$ with $A_{\mathfrak{p}}$. Since $x$ is regular, this is a regular local ring by [F5], and therefore a domain by [F6]. Put $S=A\setminus \mathfrak{p}$; by [F15], $A_{\mathfrak{p}}=S^{-1}A$. These are pointwise choices of one chart and its corresponding prime; no family of charts is chosen.

Let $C$ be any irreducible component of $X$ containing $x$. By [F7], $C$ is closed and irreducible. The subset $C\cap U$ is a nonempty open subset of $C$, so [F8] makes it dense in $C$ and [F13] makes it irreducible. It is closed in $U$ because $C$ is closed in $X$. To see it is maximal irreducible in $U$, let $Z$ be an irreducible closed subset of $U$ containing $C\cap U$. Its closure $\bar{Z}$ in $X$ is irreducible by [F9]. Since $C\cap U\subseteq Z\subseteq \bar{Z}$ and $C\cap U$ is dense in $C$, we have $C\subseteq \bar{Z}$. The maximality of $C$ then gives $C=\bar{Z}$. As $Z$ is closed in $U$, $\bar{Z}\cap U=Z$, so $C\cap U=Z$. Thus $C\cap U$ is an irreducible component of $U$. By [F10], there is a unique minimal prime ${\mathfrak{q}}_C$ of $A$ with $C\cap U=V({\mathfrak{q}}_C)$. Since $x$ corresponds to $\mathfrak{p}$ and lies in this vanishing set, [F12] gives ${\mathfrak{q}}_C\subseteq \mathfrak{p}$.

The prime correspondence [F11] identifies the primes of $A_{\mathfrak{p}}$ with primes of $A$ contained in $\mathfrak{p}$. Since ${\mathfrak{q}}_C$ is minimal in $A$, its extension ${\mathfrak{q}}_C{A}_{\mathfrak{p}}$ is minimal in $A_{\mathfrak{p}}$: a prime properly below it would contract to a prime properly below ${\mathfrak{q}}_C$. But $A_{\mathfrak{p}}$ is a domain by step 1.1, so its only minimal prime is $(0)$. Hence ${\mathfrak{q}}_C{A}_{\mathfrak{p}}=(0)$. The localization correspondence is one-to-one, so all components $C$ through $x$ have the same prime ${\mathfrak{q}}_C$. Their intersections with $U$ are therefore the same; each such intersection is dense in its component by [F8], so taking its closure in $X$ recovers that component. Thus there is at most one component through $x$.

Under the assumed AC [F1], [F14] gives at least one irreducible component through $x$. Together with step 2.1 this proves there is exactly one. If $X$ is empty, there is no point $x$ and the assertion is vacuous. If the local dimension is zero, $A_{\mathfrak{p}}$ is a zero-dimensional local domain and hence a field; the same minimal-prime argument still gives one component. No dimension restriction was used in steps 1.1–2.1. AC is also used through [F6] and [F10] for the local-domain theorem and the affine minimal-prime correspondence. For the fixed point $x$, the proof chooses one chart and makes no simultaneous choices. The statement is a uniqueness-and-existence claim, not an iff criterion; no endpoint parameter is present.