The question is how to understand why (A → B) v (B → A) is always true even when A and B are sentences that have nothing to do with each other. The reason is that the truth values for the conditional (→) and the disjunction (v) are defined to be true for three out of the four possible sets of valuations of A and B in such a way that their combination always yields a true result.
Consider the truth table for the conditional:
Note that there is only one way for the conditional to be false, but three ways for it to be true.
Similarly consider the disjunction:
Again, there is only one way for the disjunction to be false.
Furthermore, if we have A → B false then B → A would be true, but the only way for the disjunction to be false is if both of those conditionals are false. The OP noticed this:
My thought is that I can prove that it is false if I show α doesn't entail β and β doesn't entail α.
Since it is not possible for both of those conditionals to be false, that makes the disjunction true for all valuations of A and B. The disjunction (A → B) v (B → A) is a tautology.
Note that the title question asks something different which may be where some of the difficulty comes from:
Do α and β entail each other?
This question perhaps asks whether (A → B) & (B → A) or A ↔ B is always True? However, A and B do not always entail each other as the following truth table shows.
Compare that result with the truth table for the disjunction:
The disjunction is a tautology because all of the valuations of the sentences A and B lead to a True result. The conjunction expressed in the title, however, is not a tautology.
Michael Rieppel. Truth Table Generator. Generated on May 9, 2019 from https://mrieppel.net/prog/truthtable.html