Bounty expired. Will gladly re-create one if a satisfactory answer is posted in the future.
Prove: Let $f$ be a continuous function on the closed unit disc with two properties:
1. $f$ is the identity on the boundary, i.e., on the unit circle. That is, if $|z| = 1$, then $f(z) = z$.
2. $f^2$ is the identity, i.e., for all $z$ in the closed unit disc, we have $f(f(z)) = z$.
Then $f$ must be the identity function.
Motivation: I came across this question for the closed unit sphere here on MathOverflow. It seemed to me like considering the two-dimensional case might be a good place to start in trying to tackle the problem there. Ultimately, the MO question was resolved using some nontrivial results. I am wondering whether there is a proof of the question here, for the closed unit disc, which uses methods that don't extend beyond those of basic point-set topology or a first course in real analysis.
Remark: Please note that I am looking for a fundamentally different proof of the proposition above, that is, not a modification of either of the responses given at the MathOverflow link into more digestable language.
In response to a comment: If you have thought up a proof and are concerned the level of presentation is too high, then I hope you will still post it as an answer.
Best Answer
As pointed out in the comments on the MO question, this follows immediately from a 1931 theorem of M.H.A. Newman, which is Theorem 2 in this paper. Notice that Theorem 2 is basically a lemma (used to prove the main theorem 1) whose proof takes two pages, and uses nothing other than definition of manifold.