Dense regular loci on every component

Setup. The field $k$ is perfect [F2] and AC is assumed [F1]. By [F4] the scheme $X$ is locally Noetherian, so the regular locus $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is defined [F5], and it is open in $X$ by [F7]. For an open subscheme $U\subseteq X$, [F6] gives ${\mathcal{O}}_{U,x}={\mathcal{O}}_{X,x}$ for every $x\in U$, so $X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap U=\{x\in U:{\mathcal{O}}_{U,x}\text{ is a regular local ring}\}=U_{\mathrm{r}\mathrm{e}\mathrm{g}},$ and if $U=\mathrm{Spec}A$ is affine then $A$ is reduced: $X$ reduced means ${\mathcal{N}}_X=0$ [F3], the restriction of the zero sheaf is zero, and on $\mathrm{Spec}A$ the reduction is $\mathrm{Spec}(A\mathrm{/}\sqrt{(0)})$, so $\sqrt{(0)}=0$ and $A$ has no nonzero nilpotent, that is, $A$ is reduced [F3]. This proves clause 1 of the statement and records the two facts used below.

Comparing a component of $X$ with a chart component. Let $Z$ be an irreducible component of $X$; it is nonempty and closed in $X$ [F10]. Choose $z\in Z$; by [F6] there is an affine open subscheme $U=\mathrm{Spec}A$ of $X$ with $z\in U$. Then $U\cap Z$ is a nonempty open subspace of $Z$, hence irreducible [F11], and dense in $Z$ [F11]; by [F10] it is contained in an irreducible component $W$ of $U$, and by [F9] we have $W=V(\mathfrak{p})$ for a minimal prime $\mathfrak{p}$. The closure $\bar{W}$ of $W$ in $X$ is irreducible and closed [F10], and it contains $U\cap Z$, whose closure in $X$ equals $Z$: indeed $U\cap Z$ is dense in $Z$, so $Z={\overline{U\cap Z}}^Z\subseteq {\overline{U\cap Z}}^X\subseteq Z$, the second inclusion because $U\cap Z\subseteq Z$ and $Z$ is closed in $X$ [F12]. Since $Z$ is a maximal irreducible subset of $X$ and $\bar{W}\supseteq Z$ is irreducible, $\bar{W}=Z$; finally $W=\bar{W}\cap U$, because $W$ is closed in $U$ [F10] and any point of $\bar{W}\cap U$ outside $W$ would have the open neighbourhood $U\setminus W$ in $X$ disjoint from $W$. Hence $W=Z\cap U\mathrm{.}$

The affine chart input. Let $U=\mathrm{Spec}A$ be a nonempty affine open subscheme of $X$, with $A$ reduced of finite type over the perfect field $k$ [step 1.1, F4]. Let $W$ be an irreducible component of $U$; by [F9] there is a minimal prime $\mathfrak{p}\subseteq A$ with $W=V(\mathfrak{p})$, and $W\mathrm{\ne }\mathrm{\varnothing }$ because $\mathfrak{p}\in V(\mathfrak{p})$ [F10]. Since $A$ is reduced, clause 3 of [F8] applies and the regular locus of $\mathrm{Spec}A$ contains a dense open subset $D$ of $W$; in particular $D\subseteq U_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap W$ [step 1.1], the set $U_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap W$ is open in $W$ because $U_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is open in $U$, and it is dense in $W$ because it contains the dense subset $D$; also $D\mathrm{\ne }\mathrm{\varnothing }$, because a dense subset of the nonempty space $W$ cannot be empty.

Density of the regular locus on every component. With $Z$, $U=\mathrm{Spec}A$ and $W=Z\cap U$ as in step 1.2, step 2.1 applied to the component $W$ of $U$ produces the dense open subset $D\subseteq U_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap W$ with $D\mathrm{\ne }\mathrm{\varnothing }$. Here $U_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap W=(X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap U)\cap (Z\cap U)=X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap (Z\cap U)\subseteq X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap Z$ [step 1.1, step 1.2], so $D$ is a nonempty subset of $X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap Z$ that is open in $W$; since $W=Z\cap U$ is open in $Z$ (as $U$ is open in $X$), $D$ is open in $Z$. Thus $X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap Z$ is a nonempty subset of $Z$ that is open in $Z$ (ostensibly open in $X$ by clause 1, hence open in $Z$), and therefore it is dense in $Z$ because $Z$ is irreducible [F11]. This proves clause 2 of the statement, including the assertion that each component contains a nonempty dense open set of regular points, namely $X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap Z$ itself.

Nonemptiness and boundaries. If $X\mathrm{\ne }\mathrm{\varnothing }$, pick a point $z\in X$; by [F10] it lies in some irreducible component $Z$, and step 3.1 gives $\mathrm{\varnothing }\mathrm{\ne }{X}_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap Z\subseteq X_{\mathrm{r}\mathrm{e}\mathrm{g}}$, so the regular locus is nonempty; this proves clause 3. If $X=\mathrm{\varnothing }$ then there are no irreducible components [F10] and clauses 2 and 3 are vacuous, while $X_{\mathrm{r}\mathrm{e}\mathrm{g}}=\mathrm{\varnothing }$ is open in $X$. Reducedness cannot be dropped: let $k$ be a perfect field [F2], let $R=k[ϵ]\mathrm{/}({ϵ}^2)$ and $X=\mathrm{Spec}R$. By [F13] the scheme $X$ has exactly one point, the prime $(ϵ)$, whose residue field is $k$, and $X$ is not reduced, while $X$ is of finite type over $k$ because $R$ is a quotient of the polynomial ring $k[ϵ]$ [F4]. Every element of $R$ has the form $a+bϵ$ with $a,b\in k$; such an element with $a\mathrm{\ne }0$ is a unit, with inverse $a^{-1}-a^{-2}bϵ$, and the elements with $a=0$ are exactly the multiples of $ϵ$, so $(ϵ)$ is the unique maximal ideal and the localization at it is $R$ itself; the stalk at the unique point is therefore ${\mathcal{O}}_{X,(ϵ)}\cong R_{(ϵ)}=R$ [F14]. The nonreducedness of $X$ means by [F3] that the coordinate ring $R$ is not reduced, so $R$ has a nonzero nilpotent element and is not a domain, a domain having no nonzero nilpotent; since a regular local ring is a domain [F15], the local ring ${\mathcal{O}}_{X,(ϵ)}\cong R$ is not regular. Hence $X_{\mathrm{r}\mathrm{e}\mathrm{g}}=\mathrm{\varnothing }$ by [F5], as $(ϵ)$ is the only point of $X$ [F13] and its local ring is not regular, while $X\mathrm{\ne }\mathrm{\varnothing }$ is irreducible [F11] with sole irreducible component $X$ itself [F10] and $X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap X=\mathrm{\varnothing }$ is not dense in $X$ [F12], so clauses 2 and 3 fail for this finite-type $k$-scheme, which is not reduced. Perfectness is used only through [F7] and [F8] and nothing is asserted for imperfect $k$. The Axiom of Choice enters through the statement [F1] and through the suppliers that assume it, namely [F7], [F8], [F9], [F10] and [F15], each cited at the step that uses it; the remaining steps use only explicit set-theoretic and ring-theoretic operations.