"Prove the following used the method of contradiction: The sum of two consecutive integers is always odd."

I thought this proof would be a straightforward direct proof. So, the contradiction would be "The sum of two integers is always even." Sparing the rigorous details: an integer n added to another integer $(n+1)$ leads to $(2n+1)$, which contradicts the statement, since $2n+1$ is the representation of an odd number.

My teacher, however, proved this with two cases. The first case: a direct proof, using my strategy above, for $n+(n+1)$. The second case, basically a similar proof to the one in the first case but now using $(n-1)+n$. This second case is what has confused me. Isn't this step a bit redundant? Is it necessary? Does it enhance the proof, or just add superfluous information to it?

I thought this proof would be a straightforward direct proof. So, the contradiction would be "The sum of two integers is always even." Sparing the rigorous details: an integer n added to another integer $(n+1)$ leads to $(2n+1)$, which contradicts the statement, since $2n+1$ is the representation of an odd number.

My teacher, however, proved this with two cases. The first case: a direct proof, using my strategy above, for $n+(n+1)$. The second case, basically a similar proof to the one in the first case but now using $(n-1)+n$. This second case is what has confused me. Isn't this step a bit redundant? Is it necessary? Does it enhance the proof, or just add superfluous information to it?