Question Statement:
Let $f:[0,1]\rightarrow [0,1]$ be continious such that $f(0)=0$ and $f(1)=1$ and $f(f(x))=x, \forall x\in [0,1]$. Prove that $f(x)=x, \forall x\in [0,1]$.
My attempt:
Consider $f(f(x))=x$. Since the RHS is injective then $f(f(x))$ must be injective. Hence $f(x)$ must be injective. Since $f(x)$ is continuous and injective, then by the Continious Inverse theorem, $f^{-1}$ is well defined as well as continuous and surjective. Since $f^{-1}$ is defined, this implies that $f$ is unique. And since $f(x)=x$ satisfies the conditions, it must be the unique solution, hence $f(x)=x$.
However, I am not sure of the bold part. I am not really sure how I would prove that.
Here is the answer for completion. Consider $f(f(x))=x$. Since the RHS is injective then $f(f(x))$ must be injective. Hence $f(x)$ must be injective. Since $f(x)$ is continuous and injective, then by the Continious Inverse theorem, $f(x)$ is monotone.
Now for the sake of contradiction assume $f(x)\neq x$. Then $\exists x$ such that $f(x)<x$ or $f(x)>x$ (otherwise $f(x)=x$ for all $x$'s). If $\exists x$ where $f(x)<x$ then $f(f(x))<f(x)$ by monotonicity and $f(x)<x$ and hence $f(f(x))<x$ which is a contradiction to $f(f(x))=x$. Similarly if $\exists x$ where $f(x)>x$ then $f(f(x))>f(x)$ by monoticity and $f(x)>x$ contradicting $f(f(x))=x$. Hence, in both cases, there is a contradiction and hence $f(x)$ must equal $x$.