You better get over hating proofs, because that's pretty much what math is.

Clearly each of the . If it were also in , we would get , which would imply that the set is not linearly independent, contradicting the fact that it is a basis for something.

Thus, for each , . Since subspaces are closed under taking linear combinations of their elements, this implies .

For the second part,

.

But since we already know that the intersection of these two subspaces is trivial, this means that the sum is actually a direct sum: .