Acidity order of 4-halophenols
I think that, this acidity order can be justified by considering a combined effect of +R effect, -I effect and intermolecular hydrogen bonding (where it is dominant).

For 4-fluorophenol, the +R effect of $\ce F$ is significant but weaker than -I effect. The electron pair on the $\ce{2p}$ orbitals of $\ce F$ can easily delocalise with the benzene ring because of the stronger $\ce{2p\pi -2p\pi}$ overlap, which makes $\ce {+R} $ effect significant . That's why it increases the electron density on the carbon near the $\ce{-OH}$ group to such an extent which decreases its ability to donate protons, greatly decreasing 
its acidity.
Considering intermolecular hydrogen bonding also, the acidity of the compound decreases further due to strongest intermolecular hydrogen bonding between the halogens.

For 4-chlorophenol, the +R effect is negligible due to much weaker $\ce{3p\pi -2p\pi}$ overlap and thus the $\ce{-I}$ effect becomes more prominent. But $\ce {Cl}$ is electronegative enough to form strong intermolecular hydrogen bonding($\ce{HO-C6H4-Cl...H-O-C6H4-Cl...)}$. That's why, it becomes difficult to remove hydrogen atoms from $\ce{-OH}$ group, decreasing its acidity.

For 4-bromophenol and 4-iodophenol, there is no strong hydrogen bonding due to much less electronegative $\ce{Br}$ and $\ce{I}$ atoms, and also +R effect is ineffective due to very weak $\ce{4p\pi -2p\pi}$ and $\ce{5p\pi -2p\pi}$ overlap. Only -I effect is acing for these two compounds in favour of acidity. As -I effect of $\ce{Br}$ is larger than $\ce{I}$, the acidity of p-bromophenol is greater than the p-iodophenol.