A smooth graph section with a genuine end has a compact inside core

By [F3], take a smooth collar $c:S\times [0,b)\to \mathrm{\Sigma }$ and restrict to $0<a<b$. Identify $W=c(S\times [0,a])$, $U=c(S\times (0,a))$, and $\mathrm{\Gamma }=c(S\times \{a\})$ with their collar coordinates. The model ${\mathbb{S}}^2\times [0,a]\subset {\mathbb{R}}^4$ is closed and bounded, so [F4] and the smooth two-sphere parameterization of $S$ make $W$ and $\mathrm{\Gamma }$ compact by [F2]. In particular $W$ is closed in $\mathrm{\Sigma }$. By [F1] and [F6], the continuous map $G^{-1}$ sends $\mathrm{\Gamma }$ to a nonempty compact set; its radius has minimum $m>1$ and finite maximum by [F4]. Inside ${\mathrm{\Sigma }}^{\circ }\setminus \mathrm{\Gamma }$, $U$ is open and closed, since its only boundary in ${\mathrm{\Sigma }}^{\circ }$ is $\mathrm{\Gamma }$. For any $s_0,s_1\in S$, transfer a sphere path from [F7] through a smooth identification ${\mathbb{S}}^2\cong S$, obtaining a path $\alpha$ from $s_0$ to $s_1$. For $u_0,u_1\in (0,a)$, the continuous path $t\mapsto c(\alpha (t),(1-t)u_0+t{u}_1)$ stays in $U$ and joins the corresponding collar points. Composing with $G^{-1}$ proves that $H=G^{-1}(U)$ is path connected and hence connected by [F7]; it is nonempty and clopen in $X=\{r>1\}\setminus G^{-1}(\mathrm{\Gamma })$. Any connected subset of $X$ meeting $H$ cannot also meet $X\setminus H$, since its intersections with these two open sets would be a separation. Consequently the component of any point of $H$ contains $H$ by maximality and cannot leave $H$, so equals $H$. Thus $G^{-1}(U)$ is a component of $X$.

Fix $s\in S$ and let $u_j\downarrow 0$. The points $c(s,u_j)$ converge to $s$ in $\mathrm{\Sigma }$. Their inverse graph coordinates cannot have a subsequence confined to a closed annulus $1+ϵ\le r\le M$: this annulus is closed bounded and hence compact by [F4], and the bounded-sequence assertion in [F4] gives a further convergent subsequence with limit in that annulus. By [F6], continuity of $G$ would give its image a limit in ${\mathrm{\Sigma }}^{\circ }$, contradicting uniqueness of limits in the Riemannian metric topology and the boundary limit $s$. The radii are bounded above by [F1], so they tend to one: otherwise a subsequence would lie in such an annulus. The inner tail $\{1<r<m\}$ is path connected: write two points as $r_0{\omega }_0$ and $r_1{\omega }_1$, choose a unit-sphere path $\beta$ between ${\omega }_0$ and ${\omega }_1$ by [F7], and use the continuous path $t\mapsto ((1-t)r_0+t{r}_1)\beta (t)$, whose radius remains in $(1,m)$. It is connected by [F7], is disjoint from $G^{-1}(\mathrm{\Gamma })$, and intersects $G^{-1}(U)$ because the preceding collar sequence has radii tending to one. The clopen separation argument of step 1.1 forces the entire tail to lie in $G^{-1}(U)$.

Choose $0<ϵ<\min (m-1,R-1)$. Step 2.1 gives $G(\{1<r<1+ϵ\})\subset U$. Hence $B_R$ is contained in $W\cup G(\{1+ϵ\le r\le R\})$, a compact set: the annulus is closed bounded in ${\mathbb{R}}^3$ by [F4], its image is compact by [F2], and [F8] gives a finite subfamily of any ambient open cover covering each compact piece. Combining these two finite subfamilies covers their union, which is compact by the converse in [F8]. The complement of $B_R$ is $G(\{r>R\})$, an open subset of the open interior, so $B_R$ is closed in $\mathrm{\Sigma }$ and is compact by [F2]. It contains a neighborhood of $S$. Indeed, otherwise choose collar points $c(s_j,u_j)$ outside $B_R$ with $0<u_j\le a\mathrm{/}(j+1)$. Their inverse radii exceed $R$ and are uniformly bounded by [F1]. By [F4], the inverse points have a subsequence converging in a closed annulus $R\le r\le M$. Also, using the smooth identification $S\cong {\mathbb{S}}^2$ and [F4], pass to a further subsequence with $s_j\to s\in S$. Collar continuity gives $c(s_j,u_j)\to s$, whereas graph continuity by [F6] gives an interior limit, the same contradiction as in step 2.1. The countable selections here are permitted by [F3].

Away from $S$, the diffeomorphism $G$ from [F6] identifies the relative boundary of $B_R$ with $r=R$, whose sphere is closed bounded in ${\mathbb{R}}^3$ and therefore has compact image by [F4] and [F2]. By [F5], $\rho$ is continuous, so $B_R\cap \{\rho \ge \eta \}$ is closed in compact $B_R$. For every $\eta >0$ it avoids $S$, because $\rho =0$ on $S$. By [F8], its compactness therefore also holds in the interior subspace topology, which agrees with the Riemannian distance topology. In particular the stated conclusion holds for sufficiently small positive $\eta$.