Here is a different approach:

We use Tutte's theorem for the proof: Tutte theorem - Wikipedia, the free encyclopedia

We make a stronger claim : For any graph G(V,E) with |V|=2n and , for any subset S of V, the number of components in V\S is at most |S|.

Proof:The claim is obvious for |S| n.

For some S' such that S' < n, let |S'| = s. Let us assume that the number of components in V / S is more than s.

Due to the lower bound on the minimum degree, the size of each component must be at least ( n + 1 - s ). So the total number of vertices in G is at least Differentiating this function with respect to s, we find that it attains only one local extremum which is a maximum point at s = . Since it evaluates to (2n + 1) at both s=1 and s=n, it cannot have a lesser value in between.

This proves the claim and the required result follows by Tutte's theorem.