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?

The correct interpretation is that a $3$-vertex subgraph induced by $\{u,v,w\}$ is allowed to have no edges, $1$ edge, or all $3$ edges - but not $2$ of the edges.
You can check that this makes the adjacent-or-equal relation between vertices transitive. It is always reflexive and symmetric. Since it is an equivalence relation, it has equivalence classes, which are your complete graph components.
