Proof:

Let $A$ be a $3\times 3$ matrix with strictly positive real entries and $S$ be the part of the sphere

$$S = \{(x, y, z)\in \mathbb{R}^3: x^2 + y^2 + z^2 = 1,\quad x \geq 0, y\geq 0, z\geq 0\}$$

Then $S$ is homeomorphic to the closed unit disc in the plane (why?) and so has the fixed point property. If ${\bf v}$ is any unit vector with non-negative entries then the entries of $A{\bf v}$ are non-negative and at-least one of the entries must be positive. Hence the map $f: S \longrightarrow S$ given by $f({\bf v}) = A{\bf v}/\Vert A{\bf v}\Vert$ is continuous. By Brouwer's fixed point theorem, $f$ has a fixed point ${\bf v}_0$ which means $A{\bf v}_0/\Vert A{\bf v}_0\Vert = {\bf v}_0$ from which we infer that $\Vert A{\bf v}_0\Vert$ is an eigen-value of $A$ and this must be positive.