Percolation - we have either 0,1 or $\infty$ infinite open clusters

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I'm currently going over the proof of the uniqueness of the infinite open cluster for the percolation model in Grimmet (Probability on Graphs, http://www.statslab.cam.ac.uk/~grg/books/USpgs-rev3.pdf, p. 92).
There he defines the quantities $S= \{x \in \mathbb{Z}^d: dist(0,x) \leq n\}$ and $N_S(0),N_S(1)$, which are the total number of infinite clusters in $\mathbb{Z}^d$ for a realization if we declare all edges in S to be closed ($N_S(0)$) respectively open ($N_S(1)$),and $M_S$ the number of infinitely open clusters intersecting S. He deduces $P_p(N_S(0)=N_S(1)=k) = 1$ (k is the number of infinite open clusters) and claims the equivalence:
"$N_S(0)=N_S(1)$ iff. S intersects at most one infinite open cluster."
And this is were I'm unsure. The implication left to right is clear to me but for the other way around: can't we have if there is just 1 infinite open cluster that it becomes disconnected by closing edges in S and obtaining at least 2 infinite open clusters, i.e. $N_S(0) > N_S(1)$?