Definition:
A diagonal of a octagon is a line segment connecting any two non-adjacent vertices.
Prove:
Every vertex of the regular octagon will produce 2 diagonals that are parallel to at least one side and 3 diagonals that are not parallel to any side.
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.