3rd iterate of a continuous function equals identity function

Pick any $x$ and look at the three numbers $x,f(x),f(f(x))$. If any two of them are equal, then they are all equal by applying $f$ and using $ f \circ f \circ f = id$, and so $x = f(x) = f(f(x))$.

So suppose they are distinct. Without loss of generality, you can assume $x$ to be between $f(x)$ and $f(f(x))$ (if it's not, pick $f(x)$ or $f(f(x))$ in place of $x$). Then, by the intermediate value theorem, there exist a number $y$ between $x$ and $f(x)$ such that $f(y) = x$. Then, $y = f(f(f(y))) = f(f(x))$, which is impossible because $f(f(x))$ cannot be between $x$ and $f(x)$ ($x$ is between the other two)

First note that $f$ must be either increasing or decreasing by the intermediate value theorem, as if $f(x)=f(y)$ then $x=f(f(f(x)))=f(f(f(y)))=y$. Suppose $f$ is increasing. Suppose we have some $a\in\mathbb R,\epsilon_1>0$ such that $f(x)>x$ for all $x\in(a,a+\epsilon_1)$. Pick some $0<\epsilon_2<\epsilon_1$ such that $f(x)\in (a,a+\epsilon_1)$ for all $x\in (a,a+\epsilon_2)$, which must exist as $\lim\limits_{\epsilon_2\to 0^+}f(a+\epsilon_2)-(a+\epsilon_2)=0$ and for $\epsilon_2<\epsilon_1$ the expression in the limit is positive. Do the same for $0<e_3<e_2$ such that $f(x)\in (a+\epsilon_2)$ for $x\in (a,a+\epsilon_3)$. Then $$a+\epsilon_3/2=f(f(f(a+\epsilon_3/2)))>f(f(a+\epsilon_3/2))>f(a+\epsilon_3/2)>a+\epsilon_3/2$$ which is a contradiction. The same can be done if we have $f(x)<x$ for $x\in (a,a+\epsilon_1)$, so $f(x)-x$ cannot be positive or negative on any open interval, hence by continuity it cannot be positive or negative anywhere in $\mathbb R$ by continuity, hence $f(x)=x$ for all $x$. To conclude, we note that $f$ cannot be decreasing, as the same procedure gives us $f(x)=-x$ for all $x$, so $f(f(f(x)))=-x\neq x$.