Undirected tree, what about G'?

Question

Please Note: This is a part of bigger problem I finished proving one case and can't prove the following.

Let G be an undirected tree, such that G' isn't connected because it includes a node that isn't alone and not connected to all other nodes (There is some nodes that it can't reach). Find the contradiction.

Note: G' is the graph which includes all nodes in G and includes edges iff those edges do not exist in G.

Answer 1

Suppose that the vertices $u$ and $v$ are non-isolated vertices in different components of $G'$; then they are certainly not adjacent in $G'$, so they must be adjacent in $G$. If there is a vertex $w$ that is not adjacent to either of them in $G$, then $u-w-v$ is a path from $u$ to $v$ in $G'$, so there cannot be such a vertex: every vertex of $G$ other than $u$ and $v$ must be adjacent to $u$ or to $v$ (but not both — why?). In other words, $G$ must look something like this:

           *          *
            \        /
         *---u------v---*
            /|       \
           * *        *

• Show that if $v$ is the only neighbor of $u$ in $G$, then $v$ is isolated in $G'$, contrary to hypothesis. Similarly, $u$ cannot be the only neighbor of $v$ in $G$.
• Show that if $u$ has a neighbor $x\ne v$ and $v$ has a neighbor $y\ne u$, then there is a path from $u$ to $v$ in $G'$.