Let $X$ and $Y$ be the partite sets of our bipartite, and let $H$ be a clique-subgraph of $\overline{G}$. Suppose that $X_H\subseteq X$ and $Y_H\subseteq Y$ be the sets of vertices occurring in $H$, thus, $|V(H)|=|X_H\cup Y_H|=\omega(\overline{G})$. Consider, as lists, the sets $L_X=X-X_H=\{v_0,\ldots,v_k\}$, and $L_Y=Y-Y_H$. Clearly, any coloring of $H$ should consume $\omega(\overline{G})$ colors. Consider such coloring of $H$. We show, however, that the list $L_X$ can be colored using only the colors assigned to $Y_H$, and, similarly, the list $L_Y$ can be colored with only those colors of $X_H$.

Each vertex in $L_X$ is non-adjacent, in $\overline{G}$, to at least one vertex in $Y_H$ (why?!). So we may proceed, in the list order, assigning to each vertex $v_i$ the color of one of its `non-adjacencies' in $Y_H$. This process should be successful at least for $v_0$, and we show that, in fact, the process shall prevail.

Suppose that $v_t\in L_X$ is the first vertex for which the assignment process is doomed. This means, exactly, that if $S_0\subseteq Y_H$ is the set of all non-adjacencies of $v_t$, then the colors of $S_0$ have been already consumed while coloring the vertices $v_0,\ldots,v_{t-1}$. Before pursuing the end :)) , we must declare some notation.\
\textbf{Notation:}
For a subset $A\subseteq Y_H$, derive the set $D(A)\subseteq L_X$ of those vertices before $v_t$ colored by the colors of $A$. Clearly, $|A|=|D(A)|$. For a vertex $v$, let $c(v)$ denote, interchangeably, the 'current' color of $v$, or a certain other vertex with the same color.\

We run a procedure of three scenarios:

Set $i=0$ and $M= S_0\cup\cdots\cup S_i\subseteq Y_H$.

**(1)** If every vertex in $D(M)$ is adjacent to all vertices in $Y_H-M$, then the subgraph, of $\overline{G}$, induced on $(X_H\cup D(M)\cup\{v_t\})\cup(Y_H-M)$ is a complete subgraph of $\overline{G}$ whose order is $\omega(\overline{G})+1$, contradiction.

**(2)** If a vertex $w_i\in D(S_i)$ is non-adjacent, in $\overline{G}$ of course, to a vertex in $Y_H$, whose color (\textbf{VIOLET}) is not yet used, then we can re-color $w_i$ in violet, and since $c(w_i)\in S_i$, $c(w_i)$ is non-adjacent to some vertex $w_{i-1}\in D(S_{i-1})$.
Then $w_{i-1}$ can abandon the color $c(w_{i-1})$ and get colored in the color $c(w_i)$. Proceeding this shift of colors, we reach a vertex $w_0\in S_0$ to which we assign the color of $c(w_1)$, after abandoning the color $c(w_0)$. Finally giving $v_t$ the color of $c(w_0)$. Thus, we overcome the situation, and proceed coloring the rest of the list $L_X$.

**(3)** If none of the above, let $S_{i+1}\subseteq Y_H$ consist of all vertices, out of $S_0\cup\cdots\cup S_i$, that are non-adjacent to any of the vertices in $D(M)$. Increase $i$ by one and back to 1.

This algorithm should always terminate with coloring the trouble vertex $v_t$