First, we claim that $G$ is connected. Suppose, on the contary, that two vertices $u$ and $v$ are in different components. Since $u$ and $v$ are not isolated, there is a vertex $u'$ adjacent to $u$ and a vertex $v'$ adjacent to $v$. Now consider the induced subgraph on $\{u,u',v,v'\}$; it contains the edges $uu'$ and $vv'$, but by assumption it must not contain only two edges, so it must contain another edge, either $uv$, $uv'$, $u'v$, or $u'v'$, which in any case violates our assumption that $u$ and $v$ are in different components. Therefore $G$ is connected.
Suppose that $G$ is not a complete graph. Let $A$ be a maximal complete subgraph of $G$, and let $u$ be an vertex of $G$ not in $A$. Without loss of generality, we can assume that $u$ is adjacent to a vertex $a \in A$. (Proof: Choose an arbitrary vertex $a\in A$; since $G$ is connected, there is a path from $a$ to $u$, and along this path let $a'$ be the last vertex in $A$, and let the next vertex along the path be $u'$. Then $u' \notin A$, so by replacing $a$ and $u$ by $a'$ and $u'$ respectively, we may assume that $u$ is adjacent to $a$.) Now let $b$ be any vertex in $A$ distinct from $a$, and consider the induced subgraph on $\{a,b,u\}$. We are assuming that the edge $au$ is in $G$, and since $a,b\in A$ and $A$ is a complete graph we also have $ab$ in $G$; but since the induced subgraph cannot contain exactly two edges, we must also have $bu$ in $G$. Thus $u$ is a adjacent to every vertex in $A$, so $A\cup\{u\}$ is a complete subgraph, contradicting the maximality of $A$. This completes the proof.
Best Answer
Hint: There are only $7$ edges missing from $K_{10}$. Each missing edge can prevent up to $28$ different 4-vertex sets from inducing a $K_4$. How many 4-vertex sets are there?