= Solution
Let $S_+^{n-1}$ be the nonnegative portion of the <unit sphere> specified in the paper. If $Ax=0$ for some $x\in S_+^{n-1}$, then this nonzero <vector> is an <eigenvector> with <eigenvalue> zero, which already proves the assertion.
Otherwise $Ax\ne0$ for every $x\in S_+^{n-1}$, and
$$
T(x)=\frac{Ax}{|Ax|}
$$
is a <continuous map> from $S_+^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a <nonnegative matrix> and every entry of $x$ is nonnegative. By the supplied <homeomorphism> $h:\mathbb B^{n-1}\to S_+^{n-1}$, the <continuous map> $h^{-1}Th$ maps the finite-dimensional <closed unit ball> into itself. The <Brouwer fixed-point theorem> supplies a <fixed point>, hence $T(x_*)=x_*$ for some $x_*\in S_+^{n-1}$. Therefore
$$
\boxed{Ax_*=\lambda x_*,\qquad x_*\ne0,\qquad\lambda=|Ax_*|>0.}
$$
Together with the zero case this proves that $A$ has a nonnegative <eigenvalue>, with a nonnegative <eigenvector>. For $n=1$, the auxiliary <closed unit ball> is the one-point space $\mathbb B^0$ and the argument remains valid. No claim of strict positivity of every entry of the <eigenvector> is needed.
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