Prove that every component is complete graph 
Let G be a simple graph in which for every three vertices
$u,v,w$ we have:

*

*between those three vertices none of them are neighbors,

*or they form triangle

*or only two of them are neighbors.

Prove that every connected component of this graph is complete graph.

First, I am not sure what does it mean that only two of them are neighbors. Does it mean that from 3 possible edges between those three vertices only one of them exists, for example edge $e=uv$?
Then, I tried to assume the opposite, that there exists compoenent H that is not complete graph, and number of vertices in H is $\ge 1$. Now if we pick two vertices $u,v$ from H(such that $u,v$ are not neighbors because H is not complete so those two vertices exist) and $w$ from component other than H, first claim would be fulfilled, and other two won't, but that is not contradiction. Where am I mistaking?

Take $v$ with maximal number set of neighbors $N$. Then, every two in $N$ must be conected, so $N_v:=N\cup \{v\}$ is complete graph. Let $N_v$ be in a component $C$.
Suppose exists $w\in C\setminus N_v$. Then there is path from $v$ to $w$ and let $u$ be first vertex on this path (going from $v$) not in $N_v$. Then $u$ is neighbor of some $s\in N_v$ which means $s$ has bigger degree than $v$. A contradiction, so $C = N_v$.
Now you repeat the process (take vertex with maximal degree in $G\setminus C$) until you are left out of vertices.
