The first answer is wrong, and the second answer is somewhat true (alcohols are more polar than carbonyls, but not for this reason), but unhelpful.
You use protonated ethanol because it is the most likely acid present.
Let me provide an analogy.
Sulfuric acid is a strong acid. When sulfuric acid is dissolved in water, it ionizes (reacts with water)to an extent that only the tiniest fraction of $\ce{H2SO4}$ is left:
$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$
In an aqueous solution, its fair to say that the concentration of hydronium cation outweighs the concentration of free sulfuric acid by quite a lot. If you were drawing an acid-catalyzed mechanism in water, would you use $\ce{H2SO4}$ or would you use $\ce{H3O+}$?
This reaction is done in ethanol as the solvent. The basicity of ethanol is pretty similar to water, and we have a lot of ethanol, so we can expect the sulfuric acid to ionize to a great extent. These acid-base equilibrium shenanigans reaction would occur regardless of whatever else we added to the mix.
$$\ce{CH3CH2OH + H2SO4 -> CH3CH2OH2+ + HSO4-}$$
Now, if we are doing an acid-catalyzed reaction in ethanol using a strong acid, which species should we use as the proton source when drawing a mechanism: $\ce{H2SO4}$ or $\ce{CH3CH2OH2+}$?
Note that this is an example of the "leveling effect". The strongest acid that can exist in your solution is the conjugate acid of your solvent, and the strongest base that can exist in your solvent is the conjugate base of your solvent. Anything stronger reacts with your solvent.