I am trying to prove $D$. I already proved $B$. I think that for $n \geq 3$, $G$ has a Hamiltonian cycle. I tried using a few theorems that use the cardinality of the independent sets and connectivity set but they don't work because the minimum degree of $G$ is $n-1$. I considered induction but I am unsure how to proceed with that. Can someone provide some guidance? thanks!
Draw a cycle graph $C=C_{2n}.$ We can write $V(C)=X\cup Y$ where $X$ and $Y$ are independent sets, $X\cap Y=\emptyset,\ |X|=|Y|=n.$ Enlarge the graph $C$ to a graph $H$ by adding edges joining all vertices in $X$ to one another, and edges joining all vertices in $Y$ to one another. The complementary graph $\overline H$ is perfect, because it is bipartite. By the Weak Perfect Graph Theorem, the graph $H$ is also perfect. Since $H$ is perfect, the chromatic number $\chi(H)$ is equal to the clique number $\omega(H).$ Assuming $n\ge3,$ it is easy to see that $\omega(H)=n,$ whence $\chi(H)=n.$ Fix a proper vertex coloring of $H$ with colors $1,2,\dots,n.$ Now label the vertices in $X$ as $x_1,x_2,\dots,x_n$ so that $x_i$ has color $i;$ likewise, label the vertices in $Y$ as $y_1,y_2,\dots,y_n$ so that $y_i$ has color $i.$ Finally, if we use this labeling to construct the graph $G_n$ described in the problem statement, then the original cycle $C$ is a Hamiltonian cycle in $G_n.$