Suppose, for contradiction, that there were only finitely many $x \in [0, 1]$ with $g(x) = 1.$ Call these numbers $x_1, x_2, \cdots, x_t,$ with $x_1$ being the largest. Call a real number $t \in [0, 1]$ tasty if and only if there exists a nonnegative integer $c$ so that $g^{(c)}(t) \in (x_1, 1).$ Observe that for tasty reals $t$ there is not nonnegative integer $c$ so that $g^{(c)}(t) = 1,$ as this would contradict the maximality of $x_1.$ Observe that the set $X$ of tasty reals is both nonempty and not $[0, 1].$ We claim that setting $A = X$ and $B = [0, 1] / X$ produces a contradiction. We need only to show the following lemma. Lemma. If $x \in [0, 1]$ is so that $g(x) \in [0, 1]$, then $x$ and $g(x)$ are either both in $A$ or both in $B.$ Proof. We need only show that $x \in A$ if and only if $g(x) \in A.$ If $g(x) \in A$, then we have that $g(x)$ is tasty. But since $g^{(i)}(g(x)) = g^{(i+1)}(x)$ for all nonnegative integers $i$, it's clear that $x$ would be tasty too. Now suppose that $x \in A.$ Suppose, for contradiction, that $g(x)$ was not tasty. Then, because $\{g^{(i)}(g(x))| i \in \mathbb{Z}_{\ge 0}\} = \{g^{(i)}(x) | i \in \mathbb{N}\}$ has no members in $(x_1, 1)$ but $\{g^{(i)}(x)| i \in \mathbb{Z}_{\ge 0}\}$ does, it's clear that $x \in (x_1, 1).$ But then $g(x) > x$ implies that $g(x) \in (x_1, 1]$ (since we're assuming $g(x) \in [0, 1]$). As $x> x_1$, we have that $g(x) \neq 1$, and hence $g(x) \in (x_1, 1)$ and $g(x)$ is actually tasty, contradiction. So the lemma is proven. $\blacksquare$ The above lemma shows that this partition of $[0, 1]$ into subsets $A$ and $B$ doesn't satisfy the property of the problem, and so we've reached a contradiction. $\square$

Assume there are finitely many(possibly none) $x \in [0,1]$ with $g(x)=1$. The goal is designing partition $A \cup B=[0,1]$ with $x \in A \implies g(x) \notin B$ and vice versa. Starting from $1$, assume $A$ contains $1$. Then $A$ should also contain the preimage of $1$. $A$ should also contain the preimage of preimage of $1$. Repeating this, $A$ should at least contain every $x$ with $g^n (x)=1$ for some non-negative integer $n$. And stop. Let $B = [0,1] \cap A^c$. Straight from the definition of $A$, $g(x)\in A \implies x \in A$ and $1\ne x \in A \implies g(x) \in A$. If $A, B$ is not empty, then $1$ can be the only element satisfying the condition. But since $g(1)>1$, $g(1)$ cannot be the element of $B$. Next, since preimage of $1$ is finite, there exists $0<b$ such that every $b<x<1$ is not in the preimage of $1$. It can be easily seen that $B$ contains the open interval $(b,1)$. Also $A$ contains at least $1$. Therefore, $A$ and $B$ is both non-empty and contradiction came. Therefore, there are infinitely many $x$ with $g(x)=1$. Remark) Think of graph consists of vertices representing real numbers in $[0,1]$ and edge between $(x, g(x))$ if $0 \leq g(x) \leq 1$. To obey the condition, the graph should be consists of single connected component and both of the proof are finding some connected component that does not cover all the vertices.

Take $A= \{ a | g^n(a)=1 , $ for some $ n\ge 0\} $ (here we define $g^0(x)=x, g^1(x)=g(x), g^2(x)= g(g(x)),...$ etc.) Clearly $A$ is nonempty cause $1\in A$ Suppose there are only finitely many $y$ such that $g(y)=1$ and let $m<1$ be the maximal one. If $x\in A$ then either $g^k(x)=1$ for $k>0$ so $x\le g^{k-1}(x) \le m$ or $x=1$. Then $[0,1]/A$ is nonempty for it contains $(m,1)$. This dissection however leads to a contradiction beacause -If $x\in A$ and $g^k(x)=1$ , $g^{k-1}(g(x))=1$ thus $g(x)\in A$ (or $x=1$). -If $x\in [0,1]/A$ then $ g^k(x) \neq 1 $ for any $k$ and the same holds for $g^k(g(x))$.

Suppose not, and say there are finitely many preimages of $1$, and suppose $M$ is the largest, so no number in the range $(M,1)$ has image $1$. Define $S$ to be the set of reals such that on repeated applications of $g$ on it, it will eventually reach $1$. Clearly $S$ can't cover all of $[0,1)$ since nothing in the range $(M,1)$ can go to $1$ and $g$ only increases a number. Consider the partition $A = S \cup \{1\}$ and $B$ be the remaining numbers, which are both nonempty by the above line. Its easy to see that there do not exist $a \in A$ such that $f(a) \in B$ or $b \in B$ such that $f(b) \in A$, a contradiction to the problem statement, so we're done. $\blacksquare$