If x is any real number? I'd choose a number more interesting than [tex]x=\pi[/tex]. Remember that the claim employs "either/or", so if both [tex]\pi - x[/tex] and [tex]x+\pi[/tex] are irrational the claim is false.
Yes. Give a counterexample to disprove the claim. It's an exclusive or because that's the language you use when it's an exclusive or. EITHER x-pi is irrational OR x+pi is irrational is what the claim is for any real x, so if they're both irrational then that isn't "either", that's "both".
The proof is incorrect on a more fundamental level: namely, it is not a proof. It gives one example, but it is claimed that this holds for any x, hence a proof must show this. Remember, you cannot prove a general case by one example.