Reducing CLIQUE to Super Connector problem I am trying to show that our problem is NP-Complete by reducing the known problem CLIQUE to our problem.
Regular CLIQUE problem:

Input: An undirected graph $G$ and a positive integer $K$.
Goal: Does $G$ have a clique of size at least $K$?

Our problem:

Input: An undirected graph $G$ and a positive integer $K$.
Goal: Is there a group of size $K$ of super connectors in $G$?

A super connector is a node with at least $C$ (i.e., a fixed number of) edges. A group $S$ of super connectors is a set of super connectors all connected to each other.
A group of super connectors is thereby a clique and a return value of TRUE from our problem will result in TRUE in the CLIQUE problem as well.
My idea to solve this is that I modify the graph by removing all nodes that aren't super connectors. We are then left with a graph where every node is super connectors and can then input the graph in the CLIQUE problem to see if there exists a clique and thereby get a solution for our problem. By modifying the graph like this both of the problems will give the same outputs with the same inputs. Am I on the right path or completely wrong?
The simplest NP-hardness reduction is as follows. Suppose we are given an instance $(G,k)$ of CLIQUE. If $k \leq c$ then we solve CLIQUE by brute force in time $O(n^c)$. Otherwise, $G$ has a $k$-clique iff it has a $k$-clique consisting of super-connectors. This is because any vertex in a $k$-clique must have degree at least $k-1 \geq c$, and so is a super-concentrator.