Since K is a division algebra, we can find a [linear operator] mapping any point on the unit sphere to any other point. The only way the unit sphere in K can have this much symmetry is if the norm on K comes from an inner product.So, here's one way to prove it.
Suppose |·| is a norm on Rn, let G be the group of linear transformations of Rn that preserves this norm, let S be the unit sphere for this norm, and suppose elements of G can be found that map any given point on S to any other given point. We wish to show that |·| comes from an inner product on Rn.
First, note that G is compact subgroup of GL(n). It's a standard fact that any compact subgroup G of GL(n) preserves some inner product on Rn - to see this, just average an arbitrary inner product with respect to Haar measure on G. So, G preserves some inner product on Rn. Call the norm associated to this inner product ||·||. Take a point with |x| = 1 and rescale the norm ||·|| so that also ||x|| = 1. Let S' be the unit sphere for ||·||.
By assumption G acts to map x to all other points in S, but we have just seen that G preserves the norm ||·||, so G can only map x to points within S'. So, S is a subset of S'. But, both S and S' are homeomorphic to the (n-1)-sphere, so S must equal S'. So, |·| must equal ||·||. So, |·| comes from an inner product.
There might be an easier way.