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Complete any fixed bipartition to $K_{a,c}$; an uncolorable assignment stays uncolorable after adding edges. Suppose such an assignment exists with $v=a+c\le25$, and among bad four-list assignments on this complete bipartite graph choose one minimizing its palette size $N$.

Every pair of palette colors occurs together in some list. Otherwise identify a pair that never cooccurs. No list shrinks, and any proper coloring from the merged lists lifts to the original lists: the preimage at each vertex is unique, and distinct merged colors have distinct original colors. This contradicts palette minimality. Pair counting and then counting cooccurrences with each fixed color give

$$\binom N2\le6v,\qquad 4v\ge N\left\lceil\frac{N-1}{3}\right\rceil. \tag{1}$$

For a palette subset $T$ of size $\ell$, one can color using $T$ on the left and its complement on the right if every left list meets $T$ and no right list is contained in $T$. A bad assignment therefore covers every such $T$ by a left list it misses or a right list it contains. Counting, with out-of-range binomial coefficients equal to zero, yields

$$a\binom{N-4}{\ell}+c\binom{N-4}{\ell-4}\ge\binom N\ell.$$

Add the inequality with $N-\ell$ in place of $\ell$ to obtain

$$v\left(\binom{N-4}{\ell}+\binom{N-4}{\ell-4}\right) \ge2\binom N\ell. \tag{2}$$

These arguments are the palette-merging and covering method of Hanson–MacGillivray–Toft, Lemmas 1–3 and Theorem 1, pp. 184–187. The required specializations have been proved above rather than left as an uninspected source premise.

If $N\le6$, partition the palette into two parts of size at most three. Every four-list meets both, giving a coloring with separate palettes on the two sides. Thus $N\ge7$. Inequality (1) and $v\le25$ give $N\le17$. In (2) put $\ell=\lfloor N/2\rfloor$. Division by its positive coefficient gives:

All $N\le16$ are excluded. At $N=17$, the incidence inequality in (1) gives $4v\ge17\cdot6=102$, so $v\ge26$ too. This contradiction proves the claim. The exact arithmetic checker recomputes this table and the incidence bounds; the counting argument is what makes those finite palette cases exhaustive.