Option A - False. Counter Example - Complete Graph
Option B- $\overline{G}$ is connected if G is not connected. True
Proof - Let G is disconnected. Suppose u and v are vertices. If (u.v) is not an edge in G, then it is an edge in $\overline{G}$, and so we have a path from u to v in $\overline{G}$. On the other hand, if (u,v) is an edge in G, then this means u and v are in the same component of G. Since G is disconnected, we can find a vertex w in a different component so that neither (u,w) nor (v,w) is edges of G. Then (u,w) and (v,w) are edges of $\overline{G}$ and hence there exist a path from u to v in $\overline{G}$.
This shows that any two vertices in $\overline{G}$ have a path (in fact a path of length one or two) between them in $\overline{G}$, so $\overline{G}$ is connected.
Option C - At least one of G and $\overline{G}$ connected.
Which means $\overline{G}$ is connected if G is not connected and vice versa.
Proof:
If $\overline{G}$ is not connected, then there exists two disjoint sets V1 and V2 such that V1 $\subset V$ and V2 $\subset V$ and for all v1∈V1 v2∈V2 we have (v1,v2) ∉ $\overline{E}$ .This means that $ \forall v1 \in V1 \ and \ v2 \in V2$ we have $(v1,v2)∈E$ Hence G is connected.
Option D - G is not connected or $\overline{G}$ is not connected.
False. Both can be connected at the same time.
Option B is a subset of option C.
So Option C is correct. At least one of G and $\overline{G}$ connected.