Combining the collar and end barriers into interior compact confinement

The radius estimate places the entire image in $\{R\le m\}$. The collar estimate places it outside $\{\rho <\eta \}$, since within that neighborhood $b=\rho <\eta$ contradicts $b(x(v))\ge 2\eta$. The chosen $\eta$ is less than the truncation threshold, so the removed collar neighborhood is unambiguously defined.

Let $C_m=\{R\le m\}$ and let $U_{\eta }=\{\rho <\eta \}$ denote the stated collar neighborhood. Under the supplied collar diffeomorphism, $U_{\eta }$ is the image of $[0,\eta )\times B$, open relative to $[0,ϵ)\times B$; hence it is open in $\bar{S}$ and contains the entire boundary $B$. Thus $K=C_m\setminus U_{\eta }$ is closed in the compact space $C_m$ and lies in $S$. The closed-subspace theorem makes $K$ compact with the topology induced from $C_m$.

The restricted charts identify the topology of $S$ with its open subspace topology in $\bar{S}$. Consequently, if $V$ is open in $S$, it is open in $\bar{S}$, and $V\cap C_m$ is open in $C_m$. For any family of open subsets of $S$ covering $K$, their traces on $C_m$ cover $K$. Compactness from step 2.1 and the ambient-cover criterion in $C_m$ give finitely many of those traces covering $K$, so the corresponding original sets also cover $K$. The same ambient-cover criterion in $S$ now proves that $K$ is compact in $S$. Together with step 1.1 this proves interior compact confinement.