I have seen a proposition that there is no norm that can be defined on the vector space $\mathbb{K}^{\infty} = \{(x_1,x_2,...):x_1, x_2, ... \in \mathbb{K}\}$ where $\mathbb{K} = \mathbb{C}$ or $\mathbb{R}$.
I guess we need to somehow show that for any norm, there is some $u, v \in \mathbb{K}^{\infty}$ such that $u$ and $v$ don't satisfy the triangle inequality, but I can't come up with a proof.
I've also tried to use contradiction, and show that if a norm satisfies the parallelogram inequality it fails one of the norm properties, but I've also had no luck.
Thanks for any help/advice!
Edit: I was wrong, there is a norm that makes it an inner-product space. The reason it is not a Hilbert space is because it cannot be made complete. I am confused about how to show this now, though!
You have made a few mistakes in the statement so I will make a guess on what you really want to prove. ( Any norm satisfies triangle inequality and there are norms and inner products on $K^{\mathbb N}$ as mechanodroid has pointed out). So I will show that no norm on $K^{\mathbb N}$ is compatible with product topology. Let $||.||$ be a norm on $K^{\mathbb N}$ which is makes the coordinate maps continuous. Consider the closed unit ball $B$ under this norm. Since the coordinate maps are linear and continuous it follows the k-th coordinates of the points of $B$ are bounded for each $k$. By Tychonoff's Theorem this implies that $B$ is relatively compact. But that would make $K^{\mathbb N}$ finite dimensional, which is a contradiction. Thus even incomplete norms do not exist on this space.