I'm dealing with a set of $n$ ODEs:
$$\dot{x} = f(x),$$
where $x = x(t) \in \mathbb{R}^n$, $f : \mathbb{R}^n \to\mathbb{R}^n$. For the particular system I'm dealing with, the following facts are known:
- $x(t) \in A$ for all $t > 0$ if $x(0) \in A,$ where $A = (0,1)^n \subset \mathbb{R}^n$ is an open hypercube;
- the boundary of $A$, let's call it $B = \partial A$, can be only reached asymptotically;
- also, $x(t) \in B$ for all $t > 0$ if $x(0) \in B;$
- finally, $x(t) \in C$ for all $t > 0$ if $x(0) \in C,$ where $C = \overline{A \cup B}$ ($C$ is the complement of the closure of $A$).
I've proven that my system has only one steady state in $A$, let's call it $x^*_{A}$, and it has $m \geq n$ steady states on the boundary $B$, let's call them $x^*_{B,1}, x^*_{B,2}, \ldots, x^*_{B,m}$. There may be also steady states in $C$. Also, I've proven that $x^*_{A}$ is asymptotically stable, while I don't know anything about the stability of the other steady states.
Are these premises sufficient to conclude that:
$$\lim_{t \to +\infty} x(t) = x^*_A ~\forall x(0) \in A,$$
and hence $x_A^*$ is a global attractor in the set $A$?
Well, to me it seems that this information is not enough to arrive at such conclusion. I think that the following system (see picture) provides a counter-example to your statement. Let $B$ be the red invariant curve and let $A$ be the interior of the circle which is bounded by $B$. Note that $A$ is homeomorphic to $(0, 1)^2$ -- I could made it exactly $(0, 1)^2$, but it's not really important here. It's obvious that both $A$ and $B$ are forward flow invariant (in fact, they are just invariant). The same applies to set $C$ in your notation. Let red equilibrium be stable focus. Also there are two equilibria on the red invariant curve: one saddle and one stable node. So, this configuration agrees with your requirements, but the behaviour of trajectories is different from what you expect. The key element is the presence of unstable limit cycle -- it separates two basins of attraction.
There is of course a question of possibility of such or similar counter-example in higher dimensions. My opinion is that the answer is "yes", but, you know, it's just my opinion and needs proof too :) I think it's possible to put some separating hyperplane in this hypercube and organize dynamics in such way that trajectories in one half of the cube would go to the equilibrium state in the interior of the cube and other trajectories would go to an equilibrium on the boundary of the cube. But that's some sort of speculation.