Without solving explicitly show that for the IVP $x'=x^{3}-x,x(0)=0.5$ the solution converges to $0$ for $t\rightarrow\infty$

The IVP $$ x'=x^3-x, \quad x(0)=\frac{1}{2}, \tag{1} $$ enjoys uniqueness, since $f(x)=x^3-x$ is continuously differentiable.

Claim. If $\varphi$ is the solution of $(1)$, then $$ 0<\varphi(t)<1 $$ Proof. If not then $\varphi(t_1)\ge 1$, for some $t_1>0$. Hence, due to Intermediate Value Theorem, there would be a $t_1\in (0,t_1]$ such that $\varphi(t_2)=1$. In that case, $\varphi$ would satisfy the IVP $$ x'=x^3-x, \quad x(t_1)=1, \tag{2} $$ But $(2)$ is satisfied by $\psi(t)\equiv1$, and since $(2)$ enjoys uniqueness, then $\varphi\equiv \psi$. Contradiction. Similarly we can arrive to a contradiction if we assume that $\varphi(t_1)\le 0$, for some $t>0$. $\Box$

So, since $\varphi(t)\in (0,1)$, for all $t$, then (1) possesses a global solution, and $$ \varphi'(t)=\varphi^3(t)-\varphi(t)<0, \quad \text{for all $t\in\mathbb R$.} $$ Thus $\varphi$ is strictly decreasing, and therefore the $\lim_{t\to\infty}\varphi(t)$ exists and $$ \lim_{t\to\infty}\varphi(t)\ge 0. $$ We shall now show that $\lim_{t\to\infty}\varphi(t)=0$. If not and $\lim_{t\to\infty}\varphi(t)=a>0,\,$ since $\varphi$ is decreasing, then $a \le 1/2=\varphi(0)$. We would then have that for all $t>0$, $$ \varphi(t)\ge a\quad\Longrightarrow\quad \varphi'(t)=\varphi^3(t)-\varphi(t)\le a^3-a=-b<0 $$ since $g(t)=t^3-t$ is decreasing in $(0,1/\sqrt{3})$. Thus $$ \varphi(t)-\varphi(0)=\int_0^t \varphi'(s)\,ds\le -bt\to -\infty $$ as $t\to\infty$. Contradiction. Thus $\lim_{t\to\infty}\varphi(t)=0$.

Similarly, we can show that $\lim_{t\to-\infty}\varphi(t)=1$.