= Solution
Choose <simple roots> $\alpha_1,\ldots,\alpha_r$. For the orbit statement one must use the <Closed dominant Weyl chamber>
$$
\boxed{\mathcal W=\{\lambda:\langle\lambda,\alpha_i^\vee\rangle\geq0\text{ for every }i\}.}
$$
Its interior, defined using strict inequalities, is the open <Weyl chamber>. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.
Choose $\rho$ strictly inside the chamber, for example the sum of the <fundamental weights>. The finite <Weyl group> orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda,\rho)$. If $(\lambda,\alpha_i)<0$ for some <simple root>, then
$$
(s_i\lambda,\rho)-(\lambda,\rho)=-\frac{2(\lambda,\alpha_i)}{(\alpha_i,\alpha_i)}(\alpha_i,\rho)>0,
$$
contradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda\in\mathcal W$. Consequently \b[some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber]. The argument in fact works for every vector in the real <weight space>.
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