Prove finding a near clique is NP-complete
I think the following reduction works (unless I missed something):

Given $G=(V,E)$ and $k$, output $G'$ which is obtained from $G$ by adding $2$ new vertices $\{x,y\}$. that are each connected to all the vertices in $V$ (but not to each other). Finally, the output is to search for a $k+2$ near clique.

Clearly this is polynomial (linear).

If there exists a $k+2$ near clique in $G'$, then if $k+1$ of its vertices are contained in $V$, then $G$ has a $k$-clique.
If less than $k+1$ vertices are from $V$, then at least $k$ of them are, since $x$ and $y$ are not connected between them. But then, the $k$ remaining vertices in $G$ must form a clique.

Conversely, if $G$ has a $k$-clique, then just add the two new vertices $x,y$ to get a $k+2$-near clique.