This was an old exam question. I think, though, that it is some sort of trick question. Isn't this simply decidable and therefore also trivially co-semidecidable?
Because I wouldn't know how to prove that it is co-semidecidable for itself. Being co-semidecidable means that there is a chance that the algorithm (that checks whether it is a palindrome or not) never stops, even if the solution is a palindrome indeed. But I don't see why this should be the case. An algorithm should always be able to check the condition and give a clear answer, right?