A 2×2 matrix with positive entries has a positive eigenvector

Idea. A vector $x_t$ on the segment $\{x\ge0,\ x_1+x_2=1\}$ is an eigenvector iff $Ax_t$ has the same direction, i.e. iff the normalised first coordinate of $Ax_t$ equals $t$. We find such a $t$ by the IVT.

Step 1: $Ax_t>0$. For $t\in[0,1]$, $x_t\ge0$ and $x_t\ne0$; since all entries of $A$ are positive, both coordinates of $y=Ax_t$ are positive. So $g(t)=y_1/(y_1+y_2)$ is well defined, continuous in $t$, and $g(t)\in(0,1)$.

Step 2: sign change. Put $h(t)=g(t)-t$. Then $h(0)=g(0)>0$ and $h(1)=g(1)-1<0$. By the IVT there exists $t_0\in(0,1)$ with $h(t_0)=0$, i.e. $g(t_0)=t_0$.

Step 3: eigenvector. With $y=Ax_{t_0}$ and $s=y_1+y_2>0$ we get $y_1=t_0s$, and then $y_2=s-y_1=(1-t_0)s$. So
$$Ax_{t_0}=s\,x_{t_0},$$
where $x_{t_0}=(t_0,1-t_0)^T$ has both coordinates in $(0,1)$ and $s>0$.

Remark. This is the 2-dimensional case of the Perron-Frobenius theorem. The same idea (a continuous self-map of the simplex $\{x\ge0,\sum x_i=1\}$ has a fixed point) proves it in dimension $n$ via Brouwer's theorem: the map $x\mapsto Ax/\sum(Ax)_i$ is such a self-map.

Lesson. Eigenvector problems can be turned into fixed point problems on a compact convex set.