Prove that the maximum number of diagonals that can be drawn in convex n-sided polygon 

My attempt:
  Let the maximum number of diagonals of such an n-sided convex polygon = D(n) 

  For example a triangle has 0 diagonals therefore D(3)=0, similarly a
  quadrilateral by definition can only have 1 diagonal so D(4)=1.

  Assume that for any n < n_0 that D(n) = n - 3 is true.   

  Any diagonal cuts a n_0-gon into 2 parts a:
        convex j-gon, and a
        convex (n0-j+2)-gon

         3≤j<n0

   Therefore:

   D(n0) = 1 + (j-3) + (n0-j-1) = n0 - 3.

 A: The proof is clearly correct, however, I think it could be made more clear; my main criticism being is that, as it stands, it seems like your proof is trying to show:

There is a set of $n-3$ non-intersecting diagonals in any convex $n$-gon.

which is not what you want to prove.
To make the proof more clear, it would be good to note how dissecting the polygon relates to the problem. In particular, I'd say that it would be clearer if we started by:

Given a set of diagonals of a convex $n$-gon such that no two intersect, and no further diagonals can be added...

And then, you choose one diagonal from this set, and divide the polygon as you do. Right now, at that step, it seems like you're choosing a random diagonal. It's also not obvious that this creates two sub-problems - it'd be good to say that, since no diagonal could cross the one we just divided by, they must be confined to one of the polygons created by dissecting along the chosen diagonal.
It'd also be good to state that every convex $n$-gon has a diagonal (for $n>3$), because otherwise the proof is defeated if we are given no diagonals to divide across.
You have entirely the right idea - just try to rephrase it as "I am proving this thing about maximal sets of non-intersecting diagonals" - be more explicit about the object you're working with.
