A Metric, which is a (2,0) tensor on a manifold $M$ that is symmetric and non-degenerate.
A Riemannian Meric is a metric that is positive definite.
I'm reading proposition 13.3 from Lee's intro to smooth manifolds.
I understand that we can prove that every smooth manifold admits Riemannian Metric using partition of unity.
I'm just wondering is it possible to use that same proof to prove that every smooth manifold admits some other metric that is not positive definite?
I don't see how the proof relies on the positive definiteness.
You cannot, because there are differential-topological obstructions to the existence: not all manifolds admit metrics of a given signature.
You can see the discussion here for example. The simplest case is that of metrics of signature $(1,n-1)$: they exist iff the tangent bundle contains a $1$-dimensional subbundle, and that occurs iff the Euler class of the manifold is zero.