It looks so. We start with inductive proof of

**Steinitz Theorem.** Let $A$ be a finite set of rays starting from the origin in $\mathbb{R}^d$. Assume that positive span of these rays is the whole $\mathbb{R}^d$. Then there exists a subset of at most $2d$ rays from $A$ with the same property.

**Proof.** Induction in $d$. Base $d=1$ is clear. Induction step. Choose minimal $k$ such that some $k$ vectors $a_1,\dots,a_k$ which generate rays from $A$ are linearly dependent with positive coefficients: $\sum t_i a_i=0$, $t_i>0$. Then $0$ lies in interior of a $(k-1)$-dimensional simplex with vertices $a_1,\dots,a_k$. Let $X={\rm span}\,(a_1,\dots,a_k)$, $\dim X=k-1$. Factor everything modulo $X$, we get a space of dimension $d-k+1$, 0 lies in an interior of a convex hull of the image of $A\setminus {\mathbb R}_+\cdot \{a_1,\dots,a_k\}$, and it suffices to use $2(d-k+1)$ rays by induction proposition, add to them rays generated by $a_1,\dots,a_k$ to get totally at most $2(d-k+1)+k=2d+(2-k)\leqslant 2d$ rays.

If on the first step it was $k>2$, we get improved bound $2d-1$ on the number of used rays. If $k=2$, then we find a pair of opposite rays. If we proceed by induction proving your statement, we may think that two rays have generators $\pm e_d$ and generators of others are partitioned onto pairs $(x_k,\alpha_d)$, $(-x_k,\beta_d)$ for some $x_1,\dots,x_{2d-2}\in \mathbb{R}^{d-1}$. If $\alpha_d\ne \beta_d$, then considering above 4 rays, which lie in a 2-plane, it is easy to see that the whole 2-plane is generated by some three of them, so safely remove one ray.

anyclause frustrates a lot of plausible coverings. $\endgroup$ – Mark Fischler Feb 27 '16 at 1:55