Where $\chi (G)$ is the chromatic number of $G$ and $\alpha (G)$ is the independence number of $G$.
This is one of the exercises in my graph theory notes, and whilst the bound seems to be fairly lenient based off a few examples that I have tested with it I am seemingly unable to construct a proof.
One idea I had was that as $\chi (G) \geq \omega (G)$ (where $\omega (G)$ is the clique number of $G$) then the problem is equivalent to showing $$|V(G)| \leq \omega (G) \alpha (G) $$ At which point I attempted to follow an argument of supposing $|V(G)| > \omega (G) \alpha (G) $ and reaching a contradiction, but was unable to. Help would be much appreciated, thanks!
It's easier to work with $\chi(G)$. The problems are not equivalent; what you were trying to prove would be sufficient but isn't necessary (in fact, as bof points out, it's not even true).
You can colour the graph with $\chi(G)$ colours. Looking at the vertices of any given colour, they form an independent set. So how many vertices can you have in total?