Prove "almost clique" is NP complete
You can reduce to this from $CLIQUE$.

Given a graph $G=(V,E)$ and $t$, construct a new graph $G^*$ by adding two new vertices $\{v_{n+1},v_{n
+2}\}$ and connecting them with all of $G$'s vertices but removing the edge $\{v_{n+1},v_{n+2}\}$, i.e. they are not neighbors in $G^*$. return $G^*$ and $t+2$.

If $G$ has a $t$ sized clique by adding it to the two vertices we get an $t+2$ almost clique in $G^*$ (by adding $\{v_{n+1},v_{n+2}\}$).

If $G^*$ has a $t+2$ almost clique we can look at three cases:

1) It contains the two vertices $\{v_{n+1},v_{n+2}\}$, then the missing edge must be $\{v_{n+1},v_{n+2}\}$ and this implies that the other $t$ vertices form a $t$ clique in $G$.

2) It contains one of the vertices $\{v_{n+1},v_{n+2}\}$, say w.l.o.g. $v_{n+1}$, then the missing edge must be inside $G$, say $e=\{u,v\}\in G$. If we remove $u$ and $v_{n+1}$ then the other $t$ vertices, which are in $G$ must form a clique of size $t$.

3) It does not contain any of the vertices $\{v_{n+1},v_{n+2}\}$, then it is clear that this group is in $G$ and must contain a clique of size $t$.

It is also clear that the reduction is in polynomial time, actually in linear time, log-space.