Let $\mathbb{Z}^d$ be the usual $d$-dimensional lattice and let $\mathbb{H}:=\mathbb{Z}^{d-1}\times Z_+$, where $Z_+:=[0,1,2,\ldots]$. If we now consider bond percolation on $\mathbb{H}$, it is a well known result that at the critical probability for $\mathbb{H}$, $p_c(\mathbb{H})$, equals the usual critical probability $p_c$ for the $d$ dimensional lattice.
Following Grimmett's Percolation textbook, we learn that for $d\geq 2$, $\theta_\mathbb{H}(p_c)=0$, where $\theta_\mathbb{H}$ is the usual probability of 0 connecting to infinity, in this case for the half-space. The conclusion of this is that if $\theta(p_c)>0$ for $\mathbb{Z}^d$ percolation, then there exists a.s. a unique infinite open cluster in $\mathbb{Z}^d$, which is a.s. partitioned into only finite clusters by ANY division of $\mathbb{Z}^d$ into two half-spaces.
My question is, how does this fall short of proving say, $\theta(p_c)=0$ for $d=3$ and so on? It's hard for me to imagine configurations of an infinite cluster even in 3-dimensions that satisfy the slicing criteria above. If I understand correctly, the cluster would need to spiral out radially in all directions and at the very least cross the $x-y$, $y-z$ and $x-z$ planes infinitely many times. It's easy to see that such a construction fails in two dimensions (and rightfully so, $\theta(p_c)=0$ for $d=2$). Where does the difficulty lay? Or rather, if we attempt to rigorously approach a proof in this direction where do we get stuck?
