I'm really lost here guys. I'd appreciate it if you could help.
Consider the differential equation
$$ \frac{dy}{dt} = f(t, y) \tag{1} $$
with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3\ ,\ t \in \mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$ \frac{dy}{dt} = f(t, y)\ ,\ y(t_0) = 1 $$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) \neq y_2(t)$ for all $t \in \mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) \implies y_2(t) < 3$ for all $t \in \mathbb{R}$.