Consider a smooth bijection $F:\mathbb{R}^2\rightarrow\mathbb{R}^2$. A trapping region is a bounded subset $Q$ of $\mathbb{R}^2$ with the properties int$Q\neq \emptyset$, $F(Q)\subseteq Q$, and $F(Q)\neq Q$. The basin of $Q$ is defined to be the set of points in $\mathbb{R^2}$ such that $F^j(x)\in$int$Q$ for some $j\in\mathbb{N}$. The objective is to show that if $Q$ is path connected, the basin of $Q$ must also be path connected. Here is my attempt:
Let $x_1,x_2\in \mbox{bas}Q$. Then, $\exists j_1,j_2\in\mathbb{N}$ such that $F^{j_1}(x_1)\in \mbox{int}Q$ and $F^{j_2}(x_2)\in\mbox{int}Q$ by definition of basin. Let $j=\mbox{max}\{j_1,j_2\}$. Then, $F^j(x_1)\in \mbox{int}Q$ and $F^j(x_2)\in\mbox{int}Q$. Since $Q$ is path connected, there is a continuous path $s:[0,1]\rightarrow Q$ with $s(0)=F^j(x_1)$ and $s(1)=F^j(x_2)$. Since $F$ is a smooth bijection its inverse is continuous and thus $F^{-j}$ is also continuous. Furthermore, $F^{-j}\circ s: [0,1]\rightarrow \mathbb{R}^2$ is continuous with $F^{-j}\circ s(0)=x_1$ and $F^{-j}\circ s(1)=x_2$ and thus a path from $x_1$ to $x_2$. Since $x_1$ and $x_2$ are arbitrary in bas$Q$, bas$Q$ is path wise connected.
A problem I am having is showing that the path defined by the composition of the inverse of $F^j$ with $s$ actually lies completely in bas$Q$. The path $s$ is only guaranteed to lie in $Q$, not necessarily int$Q$. Then, could it be that there is a point in the path $s$ which is in $Q$ but not bas$Q$?. Such a point would be in the path $F^{-j}\circ s$ and disqualify it as a connecting path.
My comment may make for a good enough answer. There are two points:
1) With your definitions, the basin of attraction may not be path connected, even if the trapping region is.
The problem is the one you pointed out in your question: the path between $x$ and $y$ may include points which are in $Q$, but which are not in the basin of attraction of $Q$.
For instance, take:
$$F (x,y) = \left(x,y^3+\left(1- \frac{\cos(x)}{2} \right)y \right).$$
Then, for fixed $x$, the point $(x,0)$ is attracting if $x \in (-\pi/2, \pi/2)+2\pi \mathbb{Z}$, neutral repulsive if $x \in \pi/2 + \pi \mathbb{Z}$, and repulsive otherwise.
Hence, $F(\mathbb{R} \times \{0\}) = \mathbb{R} \times \{0\})$, and you can find $\varepsilon > 0$ such that $F(\overline{B}(0, \varepsilon)) \subset B(0, \varepsilon) \cup \{(\pm \varepsilon,0)\}$. Now, take:
$$Q := \overline{B}(0, \varepsilon) \cup \overline{B}(2 \pi, \varepsilon) \cup ([0, 2\pi]\times \{0\}).$$
This set has a barbell shape. You can check that, by your definition, it is a trapping region. Its interior is $B(0, \varepsilon) \cup B(2\pi, \varepsilon)$. Hence, the basin of attraction of $Q$ is:
$$\{(x,y) : x \in (- \varepsilon, \varepsilon) + \{0, 2\pi\}, \lim_{n \to + \infty} F^n (x,y) = (x,0) \}.$$
In particular, is has two connected components, separated by a distance of $2(\pi-\varepsilon)$. It is not path-connected.
2) There are better definitions of a trapping region.
We may alter the definitions so that the basin of attraction has to be path-connected. For instance, we may require that $bas(Q) = \bigcup_{n \geq 0} F^{-n} (Q)$, or that $Q$ be open - both these tweaks would imply that $bas(Q)$ is path connected. However, these modifications would be non standard. A quick look at Wikipedia suggests that the problem comes from the definition of a trapping region, the proper definition being that:
This gives us two more properties to play with. Then, we can show that if $Q$ is path-connected, then so is $bas(Q)$.
Let $U := int(Q)$. Since $Q$ is compact and $F(Q) \subset U$, there exists $\varepsilon > 0$ such that $d(Q,U^c) \geq \varepsilon$. Then, $Q' := F(Q)+B(0,\varepsilon) \subset U$ and $Q'$ is open and path-connected. In addition, given $x$ and $y$ in $bas(Q)$, there exists $j$ such that $F^j (x)$ and $F^j (y)$ are in $Q$, so that $F^{j+1} (x)$ and $F^{j+1} (y)$ are in $Q'$.
Now, take a path between $F^{j+1} (x)$ and $F^{j+1} (y)$ in $Q'$, and apply $F^{-(j+1)}$. You get a path between $x$ and $y$ which lies entirely in $bas(Q)$.