Seating $2n$ people around a table with each person has at most $n-1$ friends I am trying to show that with the setting in the title, that it is always possible to arrange the seats so that no person sits beside his/her friend.

I am not good at this kinds of problems at all, the only method I could think of is to use induction, but with the induction hypothesis (statement holds for $2(n-1)$ people) the situation seems complicated when I have $2n$ people to arrange as I could not easily single out $2$ person, any suggestions?
This question can be answered using Ore's Theorem, which states that if $G$ is a (finite, simple) graph with $n\ge3$ vertices, and if $$\deg v+\deg w\ge n$$ for every pair of non-adjacent vertices $v$ and $w$, then $G$ is Hamiltonian, that is, there is a cycle in the graph that contains every vertex in the graph. The Wikipedia essay contains a short proof of Ore's Theorem. 

The Wikipedia essay also notes that Ore's Theorem is a strengthening of Dirac's Theorem, which states that if each vertex of an $n$-vertex graph $G$ has degree at least $n/2$, then $G$ is Hamiltonian. Dirac's Theorem will suffice for solving the question here.
Construct the graph as follows:
Assume that the people are vertices and there is an edge between two vertices if and only if they are friends. Then at most $n-1$ enemies is the same as having at least $n$ friends, i.e degree of each vertex is at least $n$. Then by Dirac's theorem there exists a Hamiltonian cycle when $n \ge 3$ (In this case it will be $2n\ge 3$).
The trivial case ($n=1$) can be constructed by inspection.