Suppose $f := R^n \to R^n$ is continuously differentiable. Given $\phi(t,y)$ is the solution to the IVP: $\dot{x} = f(x), \ x(0) = y$. Assume $x_0$ is an asymptotically stable equilibrium of the system $\dot{x} = f(x)$. Show that $D = \left\{y\in R^n: \lim_{t\rightarrow \infty} \phi(t,y) = x_0\right\}$ is open and connected (an open set $D$ in $R^n$ is connected if and only if for any two points $a$ and $b\in D$ such that there is a path between $a$ and $b$ lying entirely on $D$)
My attempt: I first show that $D$ is positively invariant (in fact, the definition of $D$, together with the fact that $x_0$ is hyperbolic equilibrium, implies $D$ is the stable manifold, so $D$ is obviously invariant by Global stable/unstable manifold theorem). Then $D$ is connected because if there exists a path that connects any two points $y_1$ and $y_2$ in $D$, but that path does not lay entirely on $D$, this implies $D$ is not positively invariant (as there exists a solution that does not stay inside $D$).
I also got that $D$ must contain the point $x_0$, but even with this, I still got stuck on showing $D$ is open (I tried to use delta-epsilon definition, but failed:P). Can anyone please help me with this part?
Fix some small open ball $D_{\varepsilon}$ around point $x_0$. Obviously, for each point $x$ from $D$ there exists a moment of time $t_x$ (not unique) such that $\lbrace \phi^t(x) \rbrace_{t > t_x}$ lies in $D_{\varepsilon}$. Then you can write $D$ as $\bigcup\limits_{t \in \mathbb{R}} \phi^t(D_\varepsilon)$. Knowing that for each $t$ mapping $\phi^{t}$ is homomorphism and it maps open sets to open sets, we get that $D$ can be written as infinite union of open subsets; thus it is open itself.
I'm not sure that I've understood your proof of path connectedness fully, but I suggest you another variant. Take any two points $x$ and $y$ from $D$. There is a moment of time $\hat{t}$ such that $\phi(\hat{t}, x) \in D_{\varepsilon}$ and $\phi(\hat{t}, y) \in D_{\varepsilon}$. Since $D_{\varepsilon}$ is path-connected, take any path $\gamma$ that connects $\phi(\hat{t}, x)$ and $\phi(\hat{t}, y)$ inside $D_{\varepsilon}$. Then path $\phi^{-\hat{t}}(\gamma)$ lies inside $D$ and connects points $x$ and $y$.