$d_{X^{\lambda}}(x,y)$ is the smallest number of balls in that component forming a sequence that connects $x$ to $y$. For a set $A$ we let $C^{\lambda}(A)$ be the union of all components of $C^{\lambda}$ that intersect $A$. The length of a curve $\gamma\subset\hyp$ will be denoted by $L(\gamma)$. In this proof $\mu=\mu_{\hyp}$ and $d=d_{\hyp\times {\mathbb R}}$.

As noted earlier, it is the case that $\lambda_u(\hyp\times
  \R)<\infty$. Suppose that $\lambda_*$ is such that there is a.s. a unique unbounded component in the Poisson Boolean model with intensity $\lambda_*$ on $\hyp\times \R$. We consider the monotone coupling of the model for all intensities below $\lambda_*$. We will show that there is some intensity below $\lambda_*$ that also a.s. produces a unique unbounded component. Denote the unbounded component at $\lambda_*$ with $C^{\lambda_*}_{\infty}$. For any $r>0$, any positive integer $n$, and any $\lambda\in(0,\lambda_*)$ we define the following three random sets: $$A_1(r):=\{z\in\hyp\times \R\,:\,S(z,r)\cap
C_{\infty}^{\lambda_*}\neq\emptyset\}$$$$A_2(r,n):=\{z\in\hyp\times
\R\,:\,\sup\{d_{X^{\lambda_*}}(s,t)\,:\,s,t\in S(z,r+1/2)\cap
C_{\infty}^{\lambda_*}\}<n\}$$$$A_3(r,n,\lambda):=\{z\in\hyp\times
\R\,:\,S(z,r+2n)\cap(X^{\lambda_*}\setminus
X^{\lambda})=\emptyset\}.$$ Then put $$A(r,n,\lambda):=A_1(r)\cap A_2(r,n)\cap
A_3(r,n,\lambda).$$ Pick $y_1,\,y_2\in {\mathbb R}$ and let $$D:=D(y_1,y_2,r,n,\lambda)=\{x\in \hyp\,:\,(x,y_1)\in
A(r,n,\lambda)\mbox{ and } (x,y_2)\in
A(r,n,\lambda)\}.$$ Then $D$ is a random set in $\hyp$ such that the law of $D$ is Isom($\hyp$)-invariant. Next we will show that we can choose the parameters $r,n$ and $\lambda$ 