prety weak statement. Take an arbitrary graph $G \subseteq K_n$. Fix any four vertices $d_1, d_2, v_1, v_2$. We show that either $d_1$ and $d_2$ are connected on $G$ or $p_1$ and $p_2$ are connected on $K_n \setminus G$. FTSOC suppose that $d_1, d_2$ are disconnected and $p_1, p_2$ are disconnected suppose not. Then if the connected components of $G$ are $G = D_1 \sqcup D_2 \sqcup \dots \sqcup D_m$ where $d_1 \in D_1, d_2 \in D_2$ and likewise for $P_1 \sqcup P_2 \sqcup \dots \sqcup P_n$. Take a $m \times n$ where $(i, j)$ is filled in if $T_{ij} \coloneq D_i \cap P_j$ is empty. Since each row and column contains a filled square it follows that there exists two filled squares $(a_1, b_1), (a_2, b_2)$ which don't share a row or a column. Then there's no connections between an element of $T_{a_1b_1}$ and $T_{a_2b_2}$, contradiction due to $D_{a_1}, D_{a_2}$ and $P_{a_1}, P_{a_2}$ being disconnected.