Commuting maps on $[0,1]$ share a fixed point

Preliminary: iterates commute. From $f\circ g=g\circ f$ one checks by induction that $f^{[n]}\circ g=g\circ f^{[n]}$ and, more generally, that $f^{[n]}\circ g^{[n]}=g^{[n]}\circ f^{[n]}$ for all $n\ge1$. We will use this in the inductive step.

Part 1. Let $\phi=f-g$. Since $\phi$ is continuous on the compact interval $[0,1]$, the extreme value theorem gives
$$\alpha=\min_{x\in[0,1]}\phi(x).$$
Because $\phi(x)>0$ for every $x$, we have $\alpha>0$, and
$$f(x)\ge g(x)+\alpha\quad\forall x\in[0,1].\tag{$\ast$}$$

*Base case $n=1$.* This is exactly $(\ast)$.

*Inductive step.* Suppose $f^{[n]}(x)\ge g^{[n]}(x)+n\alpha$ for all $x\in[0,1]$. Fix $x\in[0,1]$. Since $f(x)\in[0,1]$, we may apply the inductive hypothesis to the point $f(x)$:
$$f^{[n]}(f(x))\ge g^{[n]}(f(x))+n\alpha.$$
The left-hand side is $f^{[n+1]}(x)$. By the preliminary commutation, $g^{[n]}(f(x))=f(g^{[n]}(x))$. Applying $(\ast)$ at the point $y=g^{[n]}(x)\in[0,1]$ gives
$$f(g^{[n]}(x))\ge g(g^{[n]}(x))+\alpha=g^{[n+1]}(x)+\alpha.$$
Combining,
$$f^{[n+1]}(x)\ge g^{[n+1]}(x)+n\alpha+\alpha=g^{[n+1]}(x)+(n+1)\alpha.$$
This completes the induction.

Part 2. Suppose, for contradiction, that $\phi(x)\neq0$ for every $x\in[0,1]$. Since $\phi$ is continuous on the interval $[0,1]$, the intermediate value theorem forces $\phi$ to have a constant strict sign on $[0,1]$.

*Case 1: $\phi>0$ on $[0,1]$.* Part 1 then gives $f^{[n]}(x)\ge g^{[n]}(x)+n\alpha$ for every $n\ge1$ and $x\in[0,1]$. In particular,
$$f^{[n]}(x)\ge n\alpha.$$
Fix any $x\in[0,1]$. Since $f$ takes its values in $[0,1]$, we have $f^{[n]}(x)\le 1$ for all $n$. Hence $n\alpha\le1$ for all $n\ge1$, which is impossible because $\alpha>0$ and $n\to+\infty$.

*Case 2: $\phi<0$ on $[0,1]$.* Exchanging the roles of $f$ and $g$ in Part 1 produces the same contradiction.

Both cases are impossible, so the assumption $\phi\neq0$ everywhere fails. By continuity, there is $c\in[0,1]$ with $\phi(c)=0$, i.e. $f(c)=g(c)$.