Let $$\dot{x}=F(x)$$ be an autonomous (i.e. it does not depend on $t$) system with $F: \mathbb{R}^n \to \mathbb{R}^m$ a regular as you want vector field.
Suppose also that $0$ is an hyperbolic equilibrium point, meaning that the Jacobian $JF_{|0}$ only have eigenvalues with non-zero real part. In this case $W^+$ and $W^-$ as defined below are manifolds.
Denote $$W^+=\{x: \|\phi^tx\| \to 0, \ t \to +\infty\}$$ $$W^- =\{x: \|\phi^tx\| \to 0, \ t \to -\infty\}$$ where $\phi$ is the associated flux which exists by the existence and uniqueness of solutions for $\dot{x}=F(x).$ Consider $$W^+ \cap W^- = \{x: \|\phi^tx\| \to 0, \ t \to \pm \infty\}.$$
Question 1: I read from a trustworthy source that in this case it must be that $W^+ \cap W^- = \{0\},$ but I am failing to thoroughly convince myself of this fact. Can you show me a proof ?
It could be that the notes were talking about local stable and unstable manifolds. Local stable and unstable manifolds are defined as the $x$ such that $\phi_t(x) \to x_0$ while remaining in a small neighborhood of the equilibrium point $x_0$ (as $t \to \pm \infty$).
These are (germs of) smooth manifolds intersecting each other transversally at $x_0$ (in the hyperbolic case). In particular, $W_{loc}^u(x_0) \cap W_{loc}^s(x_0)=\{x_0\}$.
More details:
Local stable and unstable manifolds are diffeomorphic to balls of dimensions $p$ and $q$, where $p$ and $q$ are the number of eigenvalues with negative/real parts in the continuous time case, and $<1$ or $>1$ modulus in the discrete case. Moreover, they are tangent to the eigenspace of these eigenvalues. Basically, the local picture is the same as the linearized vector field/discrete map.