An open, limited and connected set $C$ under the hipotesis of the theorem of inverse function with $f(\partial C)\cap C = \emptyset$ PROBLEM:
Consider $(V,||·||_{V})$ Banach, $U\subset V$ open and $f:U\rightarrow V$ of class $C^{1}$ with Frechet derivative invertible in all $U$. Supose that $A\subset U$ is open, connected and $\bar{A}\subset U$ is compact. Also suppose that $f(\partial A)\cap A = \emptyset$. Show that $f(A)\cap A=\emptyset$ or $A\subset f(A)$.
Hint: Show that $f(A)\cap A$ is open and closed in the topology induced by A
I use the fact that "if the Frechet derivative is invertible in all points and $f$ is $C^{1}$ then $f$ is an open map" to prove that $f(A)\cap A$ is open.
Now, I want to prove that $f(A)\cap A$ is closed by taking $\{p_n\}_{n\in\mathbb{N}}\in f(A)\cap A$ such that $p_n\rightarrow p$ and showing that $p \in f(A)\cap A$. We know that by definition $p\in\bar{A} = \partial A \cup A$.
Case 1: $p\in A$: As $p_n\in f(A)$, there exists $\{q_n\}_{n\in\mathbb{N}}$ such that $p_n = f(q_n)$. As $f(q_n)=p_n\rightarrow p$ and $\bar A$ is compact (limited and closed) then there exists $q\in A$ such that $q_n\rightarrow q$ and by continuity of $f$, $f(q)=p$. Now, $p=f(q)\in f(\bar A)$ but $f(q)\notin f(\partial A)$ (otherwise $p\notin A$) so $p\in f(A)$. Thus, $p\in f(A)\cap A$.
Case 2: $p\in\partial A$: Then $f(p)\in f(\partial A)$ so $f(p)\in A^{c}$. But, $f(p_{n})\rightarrow f(p)$.
That was the demonstration that was given to me, but I don't understand why in the Case 2, that is enough to hold. Can someone give me a hint?

This proof needs to be corrected. Observe that you are supposed to prove that $f(A)\cap A$ is closed in the topology induced by $A$. Therefore, if you choose $p_n\in f(A)\cap A$ such that $p_n\to p$ then you should assume that $p\in A$ (since $A$ is your metric/topological space). Therefore the second case is not needed.
Now here's my another simpler proof:
$$f(\overline A)\cap A = 
f(A\cup \partial A)\cap A =
(f(A)\cup f(\partial A))\cap A = 
(f(A)\cap A)\cup (f(\partial A)\cap A) =
f(A)\cap A.  
$$
Since $\overline A$ is compact, the set $f(\overline A)$ is also compact, hence closed. Therefore $f(\overline A)\cap A$ is closed in the topology induced by $A$.
