To see that \textsc {Independent-Set} is in NP, note that we can guess a subset of $k$ vertices in $G$ and confirm that no pair of vertices in this set are adjacent. \par To show \textsc {Independent-Set} is NP-hard, we will reduce \textsc {Vertex-Cover} to this problem. Specifically, given an instance $G,k$ to \textsc {Vertex-Cover}, we note that $G$ has a vertex cover of size $k$ if and only if it has an independent set of size $n-k$, where $n$ is the number of vertices in $G$.