Yes. The set of fixed points of $g$ is closed, nonempty, and is mapped into itself by $f$. Letting $a\le b$ be, respectively, the minimum and maximum fixed points of $g$, we have $f(a)\ge a=g(a)$ and $f(b)\le b=g(b)$. So, by the intermediate value theorem, there is an $x\in[a,b]$ with $f(x)=g(x)$.

Also, to reiterate the points made in the comments, this is a difficult problem for more general domains. The case of commuting maps on the closed disc has been asked before, and is still open. In fact, even the case of commuting maps on the simple triod (i.e., a capital 'T') appears to be an open problem, according to the contributed problem from Jeff Norden here (Commuting, coincidence-point-free maps on a triod).

Yes. The set of fixed points of $g$ is closed, nonempty, and is mapped into itself by $f$. Letting $a\le b$ be, respectively, the minimum and maximum fixed points of $g$, we have $f(a)\ge a=g(a)$ and $f(b)\le b=g(b)$. So, by the intermediate value theorem, there is an $x\in[a,b]$ with $f(x)=g(x)$.

Also, to reiterate the points made in the comments, this is a difficult problem for more general domains. The case of commuting maps on the closed disc has been asked before, and is still open. In fact, even the case of commuting maps on the simple triod (i.e., a capital 'T') appears to be an open problem, according to the contributed problem from Jeff Norden here (Commuting, coincidence-point-free maps on a triod).