The following is Exercise 11.8. from the book Differential Equations by Marcelo Viana and José Espinar.
Exercise. In the context of the Poincaré-Bendixson Theorem,show that, given two distinct stationary points in $\omega(p)$, there exists at most one regular trajectory $\gamma \subset \omega(p)$ connecting these points, that is, such that $\alpha(\gamma)$ is one of the points and $\omega(\gamma)$ is the other.
Assuming we have two curves $\gamma_1, \gamma_2$, we can get a contradiction in the case that $\alpha(\gamma_1) = \alpha(\gamma_2)$ and $\omega(\gamma_1) = \omega(\gamma_2)$. (I will put the proof below.) However, my proof does not work in the case that $\alpha(\gamma_1) = \omega(\gamma_2)$ and $\alpha(\gamma_2) = \omega(\gamma_1)$. Rereading the question, I am actually not sure if that is even part of the question.
Does the claim still hold if the $\alpha$ and $\omega$ limit sets of $\gamma_1, \gamma_2$ are "mixed"? If not, is there a (simple) counterexample?
Proof. Let $a,b \in \omega(p)$ be two distinct stationary points and assume towards a contradiction that there are two regular trajectories $\gamma_1, \gamma_2$ connecting them, with $a$ and $b$ being $\omega$ or $\alpha$-limits of the trajectories. Note that $\gamma_1\cup\gamma_2\cup \{a\} \cup \{b\}$ defines a Jordan curve, so it divides the plane into two regions, the inside $D_1$ and the outside $D_2$. Clearly, $p$ cannot be contained in the Jordan curve itself, since then $\omega(p)$ could not contain two stationary points. Let us differentiate the two cases where $p \in D_1$ and $p \in D_2$.
Suppose $p \in D_1$ (the inside) and let $x_1 \in \gamma_1$ and $x_2 \in \gamma_2$, with transverse sections $S_1$ and $S_2$ respectively. Let $y_1$ be the first time that the trajectory $\gamma_p$ of $p$ hits $S_1$, and $y_2$ be the first time it hits $S_2$. Clearly, these points exist since $\gamma_i \subseteq \omega(p)$ for $i \in \{1,2\}$. By shrinking $S_2$ if necessary, we can further assume that $\gamma_p$ hits $S_1$ first and we may also assume that $S_2$ is hit before $S_1$ is hit again. We can define a Jordan curve $J$ as in the figure below.
Note that after hitting $S_2$, the trajectory cannot leave the inside of $J$. Indeed, it cannot cross the trajectory of $\gamma_p$ or $\omega(p)$ due to uniqueness of solutions. Furthermore, at the transverse sections $S_i$ the flow points to the inside of $J$, so here it is also impossible for the flow to escape. But then, it is for instance impossible for $a$ to be in the $\omega$-limit set of $p$, a contradiction.
The case where $p \in D_2$ lies outside can be solved in an analogous way. $$\tag*{$\blacksquare$}$$