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. 
 A: 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.
A: Warning: flawed argument below. (is it better to delete it? dunno how it works here)
As already pointed out by User Some Number in his answer, $f$ is clearly a self-homeomorphism of the unit disc $D$. By contradiction, assume $f$ is not the identity. Hence there are two distinct points $p$ and $q$ (necessarily in the interior of $D$) which get exchanged, i.e. $f(p)=q$ and $f(q)=p$ (this is because $f(f(z))=z$ for all $z\in D$). Consider the straight line segment $L$ in $D$ passing through $p$ and $q$ (cutting the disc into two parts), as well as its image $fL$ under $f$. They are both homeomorphic to the interval $[0,1]\subset\mathbb{R}$.
Now choose one of the endpoints of $L$ as your starting point (call it $x_0$), say the one which is closer to $p$ than to $q$. While moving along $L$ look at the corresponding image points on $fL$. Since $x_0$ is kept fixed by $f$, both $L$ and $fL$ start at $x_0$. After some time you meet $p\in L$, and the corresponding image on $fL$ must meet $q$ because $f(p)=q$. Later on you meet $q$, and the corresponding image on $fL$ must meet $p$ because $f(q)=p$. Now if you plot the map $f\colon L\to fL$ on a coordinate system you recognize that it can't be strictly increasing (or decreasing) because the graph must pass through the points $(x_0,x_0)$, $(p,q)$, $(q,p)$, $(x_1,x_1)$. In other words, it is not a homeomorphism (every homeomorphism between intervals either preserves or reverses the order).
I guess there is a more elegant (rigorous?) way to phrase this construction (maybe transporting the order of $[0,1]$ on $L$ and $fL$ via their respective homeomorphisms).
[Notice that this argument carries over to any dimension, i.e closed balls in $\mathbb{R}^n$]
A: I apologize for not making this a comment, but I lack the reputation. Emanuele Paolini's answer is incorrect. $(x_1, 0)$ need not be under the curve $\gamma$ (since it can "dip" below). Presumably one can choose the point more carefully so that the argument works, but it doesn't seem clear since $\gamma$ can oscillate wildly.
A: I am really not sure if this works but i thought I try. The picture in the end might help to understand my idea or even to correct it.
Let $f$ be as above i.e the identity on the boundary and a bijective. Assume for contradiction $f$ is not the identity. This means that the set $A:=\{x\in D^2:f(x)\neq x\}$ is not empty and is contained in the interior of $D^2$. 
Now consider some  injective and straight $\phi_1:[0,1]\rightarrow D^2$ s.t $\phi(0)=x$ with $||x||=1$ i.e the "path" $\phi_1$ starts on the boundary and ends $\phi_1(1)=-x$ on the antipodal of $x$. Morever choose $\phi_1$ s.t $\phi_1([0,1])\cap A\neq\emptyset$. Just imagine a stright line going through the origin and hitting some element in A.
Consider now an arbitrary $x_0\in A$.Then $f(x_0)$ can't be an in $\phi_1([0,1])$ since otherwiese $(f\circ\phi_1)([0,1])$ would not be path connected any more as a consequence of continuity of $f$ and that $\phi(0)$ and $\phi(1)$ are fixed points of $f$. This yields that $f(x_0)$ lies somwhere in the interior of $D^2$ but not on the path $\phi_1$. Furthermore $(f\circ\phi_1):[0,1]\rightarrow D^2$ is a new path in $D^2$ which has the same start and end points but is different of $\phi$ at least for some small epsilon neighborhood of $x_0$. Finally consider $\phi_2:[0,1]\rightarrow D^2$ which is a straight line through $x_0$ and $f(x_0)$ starting and ending on the boundary and assume wlog that it hits $f(x_0)$ first. 
Then $(f\circ\phi_2)([0,1])$ contains $x_0$ since $\exists t \in [0,1]$ s.t $f(\phi(t))=f(f(x_0))=x_0$ but does not contain $f(x_0)$ anymore by the same argumentation as above. Moreover since it start on the boundary end ends on the antipodal point on the boundary, $(f\circ\phi_2)([0,1])$ cuts $(f\circ\phi_1)([0,1])$ in some additional point $z_0$. This implies that there exists some $y_1\in\phi_1([0,1])$ and some $y_2\in\phi_2([0,1]) $ s.t $f(y_1)=f(y_2)=z_0$ for $y_1\neq y_2$ what contradicts that $f$ is bijective...

A: 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 $f(z)=z$ 
