Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?

Let $x'=f(t,x)$ be a differential equation with $f$ in the hypothesis of Picard's theorem. Let $\varphi$ be a solution such that its interval of definition contains $(t_0,+\infty)$ for some fixed $t_0\in \mathbb{R}$. Suppose $\lim_{t\to+\infty} \varphi(t)=a \in \mathbb{R}$. Must the constant function $t\mapsto a$ be a solution of the equation?

(This is a question I've asked myself while studying the logistic model $x'=ax(1-x)$ and wondering how to justify the solutions look the way they do.)

EDIT: By "the hypothesis of Picard's theorem", I mean $f$ continuous and locally Lipschitz with respect to $x$.

I'm interested also in particular cases (e.g. autonomous system) where this holds.

You have not been explicit about "the hypotheses of Picard's theorem" (and yes, different sources state these differently) but in any case I believe that the answer is no unless you are far more restrictive on what you require of $f$.

Consider $f$ defined on $(1, \infty) \times \mathbb{R}$ by $$f(t,x) = \frac{\cos(t) - x}{t}, \qquad t > 1, \quad x \in \mathbb{R}.$$ Whatever your hypotheses are I expect they are satisfied by this $f$ on $(1, \infty) \times \mathbb{R}$.

Consider $y: (1, \infty) \to \mathbb{R}$ defined by $$y(t) = \frac{\sin(t)}{t}, \qquad t > 1.$$ A short calculation (just the quotient rule) shows that $$y'(t) = \frac{t \cos(t) - \sin(t)}{t^2} = \frac{\cos(t) - \frac{\sin(t)}{t}}{t} = \frac{\cos(t) - y(t)}{t} = f(t,y(t)), \qquad t > 1,$$ and clearly $$\lim_{t \to \infty} y(t) = 0.$$ Yet $f(t,0) = \frac{\cos t}{t}$ is not identically zero, so the constant function $0$ is not a solution to $x' = f(t,x)$.

This is an elaboration of my previuous comment, written at Bruno's request.

Consider the equation $x'=f(x)$ with $f\colon\mathbb{R}\to\mathbb{R}$ continuous and let $x\colon(t_0,\infty)\to\mathbb{R}$ be a solution such that $\lim_{t\to\infty}x(t)=a$. I claim that $f(a)=0$, and hence $z(t)\equiv a$ is a constant solution. The proof is by contradiction.

Suppose $f(a)\ne0$. Without loss of generality we may assume that $f(a)>0$. We have $$\lim_{t\to\infty}x'(t)=\lim_{t\to\infty}f(x(t))=f(a).$$ Let $\delta=f(a)/2$. There exists $t_1\ge t_0$ such that $x'(t)\ge\delta$ if $t\ge t_1$. In particular, $$x(t)=x(t_1)+\int_{t_1}^tx'(t)\,dt\ge x(t_1)+(t-t_1)\delta\quad\forall t\ge t_1,$$ which contradicts the fact that $\lim_{t\to\infty}x(t)=a$.

The proof carries over to autonomous systems of equations.