 Solution

Fix a proper colouring with exactly $k$ nonempty colour classes. For any class $C_i$, suppose every vertex in it misses at least one of the other $k-1$ colours in its neighbourhood. Recolour each of these vertices by one of its missing colours. No edge joins two vertices of $C_i$, so these simultaneous recolourings cannot conflict with one another, and the chosen colours do not conflict with vertices outside $C_i$.

This would eliminate colour $i$ and produce a $(k-1)$-colouring, contradicting optimality. Therefore each colour class contains a vertex adjacent to at least one vertex of every other colour. That vertex has degree at least $k-1$. Choosing one from each class gives $k$ distinct vertices, provingFor $k=1$, a nonempty graph already has a vertex of degree at least zero, so the conclusion holds as well.