First, note that regardless of approach, we need only investigate diagonals $1$, $2$, and $3$ because there is reflective symmetry over diagonal $3$. Moreover, we only need to prove this fact for a single vertex due to rotational symmetry.

Now this probably isn't the most elegant, but one method would be to inscribe the octagon within a unit square:

$\qquad \qquad \qquad \qquad$

We can imagine this square has its lower left vertex at the origin, and we can find the coordinates of the vertices of the octagon since the lengths $x$ and $y$ obey the following:

\begin{align} &2x + y \ = \ 1 \qquad \ \quad \text{(It's a unit square)} \\ &2x^2 \ = \ y^2 \qquad \qquad \text{(From the Pythagorean theorem)} \end{align}

With those coordinates, one can explicitly compute and compare the slopes of the diagonals and of the sides.