Proving a certain map on the closed unit disc must be the identity 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.
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.
Consider the fact that $f(f(z))=z$. This shows that $f$ is its own inverse. Since $f$ is invertible, it is a bijection; a bijection from the compact space closed disk into the Hausdorff  space of the closed disk, so that it is a homeomorphism (continuous bijection from compact to hausdorff is a homeomorphism). Since the closed unit disk is contractible, any two maps into it are homotopic, so $f$ is homotopic to the identity, so that , from the conditions on the boundary,$f(z)=z$ or $f(z)=-z$. From the conditions given, we must have $f(z)=z$