Perhaps I wasn't clear in my answer, but I was pointing out that just for the existence of embeddings, this follows simply from compactness arguments -- it is P

Perhaps I wasn't clear in my answer, but I was pointing out that just for the existence of embeddings, this follows simply from compactness arguments -- it is Proposition 3.4 in the paper I cited (not even a theorem). However this dimension depends on the finite subcover which in turn depends on (M, g). That is a completely different question that trying to find a dimension that is independent of (M, g) and only depends on the dimension of M rather that it's geometry. The AMS article you cited starts off only discussing this second question. So again, if you just want to know "can I embed a manifold", the answer is "of course you can". But if you ask, "Can I embed every manifold of dimension n into a space of dimension f(n)", then that is a much harder question, deserving of a person's name and the status of "theorem". The first example of the paper is a conjecture that f(n) = n(n+1)/2. But if you are just asking, "who showed you can embed a given manifold into a given dimension isometrically", then I maintain this follows rather simply from partitions of unity and finite subcovers, and was known pretty much at the same time as the definition of Riemannian manifold was formalized.