Why can we not find n+ 2 vectors in $R^n$ so that the dot product of any two of them is negative? Take for example 3-space, where you can arrange 4 vectors so that the dot product of any two is negative, but when you add a fifth vector there is a least one non-negative dot-product. 
I have seen examples of proofs using induction but they are too advanced for my background in linear algebra (very new). 
Is there a way that this can be proved using angles for the example above, and also more generally.

It might be better to just bite the bullet and try to understand the inductive proof.  It's much simpler (and, once you understand it, more intuitive) than any proof you will get based on angles.  In outline, in the inductive proof, you basically show that you can fix one of the vectors $x$ and sequentially modify another $n-1$ (by adding positive multiples of $x$ and of the vectors that have already been modified) to create an orthogonal basis such that these still have negative dot products with the other unmodified original vectors.  If there are still two other vectors left, each having negative dot products with an orthogonal (and hence with the related orthonormal) basis, they each have all negative coordinates when written with respect to that basis and so much have a positive dot product with each other.
The induction is used to show that you can construct the basis.  Without the inductive step, you'd have a hard time showing that the original set of vectors span the space.
