Suppose we have an autonomous first-order ordinary differential equation
$$\frac{dx}{dt} = f(x) \tag{*}$$
where $f$ is continuously differentiable for all $x \in D \subseteq \mathbb R$ s.t. the ODE has a unique maximal solution for every initial value problem in $D$.
Book says
Theorem: Let u be a solution s.t. $|u(t)| \le K$, for some $K > 0$ for $t \ge t_0 \color{blue}{\in I \subseteq \mathbb R}$. Then $\lim_{t \to \infty} u(t)$ exists and $f(\lim_{t \to \infty} u(t)) = 0$
Note: $\color{blue}{\text{added by me. not sure though.}}$
The intuition given is that if a solution remains $\color{green}{\text{bounded above}}$, it must be asymptotic to a constant solution.
After the proof of that, book says:
Using a similar argument, we can show that if $|u(t)| \le K$, for some $K > 0$ for $t \color{red}{\le} t_0 \color{blue}{\in I \subseteq \mathbb R}$, then $\lim_{t \to \color{red}{-\infty}} u(t)$ exists and $f(\lim_{t \to \color{red}{-\infty}} u(t)) = 0$
The intuition for this part is that a solution that is $\color{green}{\text{bounded on the left}}$ will approach a constant solution as $t \to -\infty$.
Do the green parts seem inconsistent?
I think the former should be $\color{green}{\text{bounded on the left}}$ (referring to the $t \ge t_0$) and the latter should be $\color{green}{\text{bounded on the right}}$ (referring to the $t \le t_0$).
Where is the upper bound for the '$\color{green}{\text{bounded above}}$' part?
You really need to write "bounded" instead of "bounded from above" and "bounded from below".
For counterexamples you can take $x'=x$ and $x'=-x$. The solutions starting respectively say at $-1$ and $1$ are bounded from above and bounded from below, but they don't have the properties in the theorem and in the corresponding result for negative time.
PS: The theorem should read "Then $I$ contains $[t_0,+\infty)$ and ..." and similarly for negative time.