Proving that Max Weighted Independent Set is in NP
The issue is with the version of the problem you are using. Note that as you define it, the output is required to be the maximum weight independent set, i.e. the optimum answer. $NP$ however is a class of decision problems, so the only valid outputs are Yes and No.

So if you want to show the problem is in $NP$ you first need to convert it to a decision problem:


  $k$-Weight Independent Set
  Input: A vertex weighted graph $G=(V,E,w)$ and an integer $k$.
  Question: Is the a set $V'\subset V$ such that $V'$ is an independent set and $\sum_{v\in V'} w(v) \geq k$?


It should be easier to see that this version is in $NP$. There is an optimization class - $NPO$ - that corresponds to $NP$, but the normal definition is that a problem is in $NPO$ if its decision variant is in $NP$ - so you're still in the position where you want to deal with the decision variant.