Proving that $\chi(G)=\omega(G)$ if the complement of G is bipartite. ***Please note I am aware that this question was already asked here: Proving that $ \chi(G) = \omega(G) $ if $ \bar{G} $ is bipartite.
however, I have a reason to believe the answer given in that topic is flawed.
What I so far tried: I divided the complement of G to parts V and U, while all isolated vertices belong to V and |V| is bigger than |U|. $|V|≤\omega(G)$ so I gave the vertices of V distinct colors (they are a clique in G), now all that is left to show is that the vertices of U can be colored in G without adding extra colors. the answer in the linked topic suggests that since every vertex of U has a neighbor in V in the complement of G, you can give each vertex in U the color of one of its neighbors. that is not really the case, since two vertices of U can potentially only be adjacent to the same vertex in V, and since U is a clique in G, you cannot give them the same color. this method can only work if one can prove that the complement of G has a matching that spawns all vertices of U. this is not the case for certain graphs, and so this method fails.
edit: here is a complement graph for which |V| is smaller than $\omega(G)$. (top two blue vertices and lower four red vertices form a 6 clique in G.)


Using the notation $G^c$ for complement, and $\alpha(G)$ for the stability number of $G$
The proposed proof went just a bit fast on the definition of $U$ and $V$. With its definition, you could have that $V$ is not the biggest independent set in $G^c$.
The key point is that you need $\vert V \vert \geq \vert U \vert$ before taking into account the isolated vertices of $G^c$, so that $\vert V\vert = \omega(G)=\alpha(G^c)$
Then if two elements of $U$, $u_1$ and $u_2$ are connected to the same vertices $v$ of $V$, then at least one of them must be connected to another vertices $v' \in V$.
The detailled written proof is a bit tedious, but try drawing appropriate graph, it's much easier to see. If need be I'll put the proof here tonight.
