Since the OP asked for a rigorous proof, here is one.
By Acemoglu's inequality in the first part of his proof, we can separate $\{x(t)\}_{t=0}^{\infty}$ into two subsequences, an increasing subsequence $\{x(t_{i})\}_{t_{i}\in I}$ bounded above by $x^{*}$ and a decreasing subsequence $\{x(t_{j})\}_{t_{j} \notin I}$ bounded below by $x^{*}$. Indeed, we define $I$ to be the set $\{t \in \mathbb{N} \mid x(t)<x^{*}\}$. By Acemoglu's inequality, $s>t$ implies $x(s)$ is strictly closer to $x^{*}$ than $x(t)$ is. Thus, since $x(t_{i})<x^{*}$ for all $t_{i} \in I$, $\{x(t_{i})\}_{t_{i}\in I}$ is increasing. Similarly, $\{x(t_{j})\}_{t_{j} \notin I}$ is decreasing and bounded below by $x^{*}$
Assume that each of these subsequences is infinite. Then each converges to a limit, which we will denote by $y$ and $z$ respectively.
Note that it also follows from Acemoglu's inequality that $y=x^{*}-\epsilon$ and $z=x^{*}+\epsilon$ for some $\epsilon \geq 0$. Indeed, suppose that this is not the case (we will show this generates a contradiction). For convenience, assume $\vert y -x^{*}\vert<\vert z-x^{*}\vert$. Define
$$\delta:=\vert z-x^{*}\vert-\vert y-x^{*}\vert>0.$$
Since $\lim_{i \to \infty}x(t_{i})=y$, we can pick $t \in \mathbb{N}$ such that
$$\vert x(t)-y\vert<\frac{\delta}{2}.$$
The triangle inequality then tells us that
$$\vert x(t)-x^{*}\vert <\vert y-x^{*}\vert+\frac{\delta}{2}.$$
Now pick some $s \in \mathbb{N}$ such that $s >t$ and $x(s)>x^{*}$ (such an $s$ exists by our assumption that both the constructed subsequences are infinite). Since $x(s)>z>x^{*}$, we have that
$$\vert x(s)-x^{*}\vert>\vert z-x^{*}\vert.$$
Thus,
$$\vert x(t)-x^{*}\vert <\vert y-x^{*}\vert+\frac{\delta}{2}<\vert z-x^{*}\vert <\vert x(s)-x^{*}\vert.$$
But since $s>t$, this contradicts Acemoglu's inequality.
Now we will consider the value of $g(y)$. Since,
$$g(y)=g(\lim_{i \to \infty}x(t_{i}))=\lim_{i \to \infty}g(x(t_{i}))=\lim_{i \to \infty}x(t_{i}+1),$$
we must have that $g(y)=y$ or $g(y)=z$.
If $g(y)=z$, then we must have that $y=z=x^{*}$ since if $y\neq x^{*}$, we get a contradiction. Indeed, if $y \neq x^{*}$, then Acemoglu's inequality implies that $g(y)$ is strictly closer to $x^{*}$ than $y$ is; but $y$ and $z$ are equidistant from $x^{*}$. Thus $g(y)=y$ and similarly $g(z)=z$. (If only one of the subsequences is infinite, we get this immediately.) But this implies that $y=z=x^{*}$, as Acemoglu's inequality allows us to infer that if $y \neq x^{*}$, then $g(y)$ is strictly closer to $x^{*}$ than $y$ is.
