Showing that only solution that attains a value of $1$ is constant solution. For the following ODE
\begin{align}
x'(t) = x(t)(1-x(t)^2), \ x(t) \geq 0 \  \forall  t,
\end{align} 
I wish to prove that if a solution $x$ to the ODE attains a value $x(t') = 1$ for some $t'$, then $x$ is the constant solution $x(t) = 1$ for all $t$. 
How could one go about showing it? Or is it not true?

I think this is true since after looking at numerical solutions (see image) with various initial conditions, it seems that all solutions are attracted asymptotically towards the constant solution $x = 1$ (other than the constant solution $x = 0$).
This makes sense, since when $0 < x(t) < 1$, we have $x'(t) > 0$, and when $x(t) > 1$, we have $x'(t) < 0$.
I am having trouble turning this intuition into a proof.

 A: For this to be true we need to assume that the domain of definition is connected (I leave it to you to convince yourself that it is not true if we drop this assumption). 
We consider the system
$$ (\star) =\begin{cases} x'(t) &= x(t) (1-x(t)^2) \\ x(t_0)&=1 \end{cases} $$
Note that the function $f(x)= x (1-x^2)$ is locally lipschitz and thus by the Picard-Lindelöf Theorem there exists a nbhd $(t_0-\varepsilon, t_0+ \varepsilon)$ of $t_0$ such that $(\star)$ admits a unique solution $x(t)$. However, $\tilde{x}(t) =1$ is a solution of $(\star)$. Thus, we get by the uniqueness $x(t) =1$ for $t\in (t_0 - \varepsilon, t_0+\varepsilon)$.
Hence, if we have a solution $x$ of $(\star)$ which is defined on $ I:=(T_{-\infty}; T_\infty)$ (of course we assume $t_0\in I$). We define
$$ S:= \{ t\in I \ : \ x(t)=1 \}.$$
As $x$ is continuous we get that $S$ is closed. On the other hand we can pick $t_1\in S$, then $x$ solves 
$$ (\star\star) =\begin{cases} x'(t) &= x(t) (1-x(t)^2) \\ x(t_1)&=1 \end{cases} $$
and by the argument above we get that $x=1$ in a nbhd of $t_1$, which implies that $S$ is also open. However, $I$ is connected and thus we get $S=I$ which means $x=1$.
A: This is a separable ODE which you should find has the variable solution
$$x(t)=\pm\frac1{\sqrt{1+Ce^{-2t}}}\qquad C\in[-1,\infty)\setminus\{0\}$$
or the constant solutions
$$x(t)=0,\pm1$$
For $x(t)\ge0$ this leaves the only possible solutions as
$$x(t)=\frac1{\sqrt{1+Ce^{-2t}}}\qquad C\in[-1,\infty)\setminus\{0\}$$
$$x(t)=0,1$$
As you can see, the variable solution cannot equal $1$ for any $t$ as $e^{-2t}\ne0$ for any $t$. So that leaves a choice from the constant solutions, of which $x(t)=1$ is the only solution that has $x(t')=1$ for some $t'\ge0$.
