NP hardness of partitioning a graph into two subgraphs
I think you can in fact do this by showing a reduction from the Partition problem, as stated by Chapichap. Recall the direction of reduction that is needed to show a problem is NP-Hard. In general, it is okay if the problem you have is harder than the problem you are using to show hardness (in this case, the partition problem). This is because the reduction would from the partition problem to the given problem. For the partition problem, if we are given a set of numbers $S$, and we want to find a partition $S_1,S_2=S\setminus S_1$, we need to show how to construct a graph $G$ and a number $W$ such that the vertices of $G$ can be partitioned as described (where the sum of weights of the vertices is at least $W$) if and only if the set $S$ can be partitioned.

Consider the following reduction: Let $G = (V,E)$ be a graph where $V=S$. That is, for each number in $S$, create a vertex in $G$. The edge set seems irrelevant here, so we can let $E=\varnothing$. The weight of each vertex is equivalent to the number in $S$ to which it corresponds. Finally, let $\lceil W=(\sum_{x\in S} x)/2\rceil$. That is, $W$ is the sum of the weights of all the elements of $S$, divided by two.

Asking for a partition in $G$ such that each partition has total weight at least $W$ means asking for each partition to contain half of the total weight of the graph. To be thorough, consider the case where $\sum_{x\in S}$ is even. Then $W$ is exactly half of the total weight of all vertices in the graph, and so for each partition to have a total weight of at least $W$, each partition must have equal weight. Thus such a partition of the vertices is equivalent to an equal-sum partition of the elements of the original set. Now, consider the case where $\sum_{x\in S}$ is odd. In such a case, $S$ has no partition such that the sum of the partitions are equal. However, the way we constructed $W$ ensures that $2W = (\sum_{x\in S})+1$ and so it is impossible to partition the graph so that each partition has total weight at least $W$ (the existence of such a partition would imply that the total weight of the graph is strictly greater than $\sum_{x\in S}$, which is impossible).

Hopefully you can see from the above that $G$ has a partition into two subgraphs such that the sum of the weights of vertices in each partition is at least $W$ if and only if the original set $S$ had a partition of its elements into two subsets such that the sums of elements in each subset are equal.