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Hence any Banach space of dimension infinite is the union of a countable set of hyperplanes. Baire theorem is no problem since an hyperplane may be of second Baire category. In fact there was a conjecture by Klee and Wilansky that first category hyperplane are the same as closed hyperplanes. I proved this conjecture is not true Dense Hyperplanes of First Category. Mathematische Annalen. Vol. 249. 1980. Pag. 111-114.
A vector space $V$ of dimension infinite is always union of a countable set of proper subspaces. Take a Hamel basis $e_\alpha$ . Each vector can be written $v = \sum_\alpha x_\alpha e_\alpha$. Let $V_\alpha$ the subspace of those vectors for which $x_\alpha = 0$ It is clear that $V$ is the union of any sequence $V_{\alpha_n}$ for $n\in \N$ with $\alpha_n$ differents.
Cutting a set = the intersection with this set is non empty. You asked for "positively separated sets". I understodd that the infimum of the distances between points one of each set is greater than a certain $d > 0$. I called this separated sets in my answer. So my answer is the same as Edgar.
Apparently I can not post here a plot. The x-ray gives little doubt that the zeros of $\Gamma(s)-\Gamma(1-s)$ are the ones with real part $1/2$ that appeared computed in this question. The complex at $-1.69-0.30 i$ its complex conjugate the symmetrical of this with respect the critical line $2.69+0.30i$ and its complex conjugate. And then the real zeros one at $0.5$. The others real zeros can be obtained best from a real plot of the function. In the x-ray this zeros, that are very near the poles at $4$, $5$, $\dots$, can not be seen since they are contained in very short lines.
