@blue, the resonance structure is correct. The second structure, the bromooxonium ion, needs to have curved arrows (electron movements) consistent with its formation. If that was the reaction with HBr, that would give H- as the leaving group. That is not consistent with the ionization of HBr.
Protonation of your resonance structure should now make sense. You then have two choices for the product, addition of the bromide to the oxygen stabilized carbocation or an SN2 opening of the ring. I don't know what might have been inferred in your class, so pick the reaction consistent with it. (I predict the bromoaldehyde is the actual product though).
@macman, granted that carbon is a better inductive electron donor than fluorine, but fluorine is a better resonance donor and resonance effects generally are stronger.
If fluoride can hydrogen bond, that simply indicates its electrons are available to hydrogen bond. That would be consistent with an increasing bond strength from oxygen to nitrogen and carbon (carbanion) with hydrogen bonds becoming covalent bonds. Hydrogen bonding effects are much harder to find in the third row because the greater nuclear charge increases the electron withdrawing character of those atoms and an increasing HX acidity. Therefore, while fluorine can hydrogen bond, hydrogen bonding virtually disappears for Cl, Br, etc.
I offered acidity, hydrogen bonding, and Hammett sigma values that show the other halogens are more electron withdrawing than fluorine. (I personally think acidity is perhaps the best method to determine heteronuclear ionization values. It is well studied, widely applied, water is virtually a universal solvent for a very large number of acids, quite consistent within the ionization range to which it can be applied.) Perhaps you could give me an example that shows fluorine more electron withdrawing than bromine (though perhaps in another thread or message as such a discussion would highjack the original posters question).