(False?) proof for computability of a function?
Think of it this way, Mike: This proof is "branching" into multiple possible cases, one of which has to be true (using the law of excluded middle that for every proposition $p$, either $p$ is true or $\neg p$ is true). But at the end of each of these branches, you always manage to prove that the function $f$ is computable. Therefore, no matter which of the cases actually holds in real life, $f$ must be computable. (However, the precise reason why $f$ is computable will be different, depending on the branch.)