= Solution
The <Lusternik-Schnirelmann-Borsuk theorem> has the following two equivalent covering formulations. For every integer $d\geq0$, a cover of the <sphere> $S^d$ by $d+1$ <closed sets> has a member containing an <antipodal pair>. The same assertion holds with <open sets> in place of <closed sets>. Thus \b[$d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$.] The dimension-zero case simply says that one set covering the two-point <sphere> contains both points.
Here is why the two versions agree. A finite <open cover> of a <compact metric space> admits a <closed> shrinking that still covers: sufficiently small <closed balls> subordinate to the <open cover> can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty <closed> subset $C$ of $S^d$ avoids <antipodal pairs>, <compactness> gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids <antipodal pairs>. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.
Another common equivalent formulation is the <Borsuk-Ulam theorem>: every <continuous> $f:S^d\to\mathbb R^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every <continuous> <odd function> $g:S^d\to\mathbb R^d$ has a zero. For example, if $d+1$ antipodal-pair-free <open sets> covered $S^d$, a subordinate <partition of unity> $(\phi_1,\ldots,\phi_{d+1})$ would give the <odd function>
$$
g(x)=(\phi_i(x)-\phi_i(-x))_{i=1}^{d+1}
\in\{y\in\mathbb R^{d+1}:\textstyle\sum_i y_i=0\}\cong\mathbb R^d.
$$
It cannot vanish, since some $\phi_i(x)>0$, whereas antipodal-pair-freeness forces $\phi_i(-x)=0$. In the other direction, if an <odd function> $g:S^d\to\mathbb R^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a <regular simplex> centred at the origin in $\mathbb R^d$. For $d\geq1$, the <open sets> $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an <antipodal pair>, contradicting the covering theorem.
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