Folland proof of proposition 4.1 screenshot of proposition from folland here
I am having trouble understanding the last part of the proof. Since $x \not\in A \cup acc(A) $, then there exists a open $U$ containing $x$ such that $U \cap A = \emptyset$. I do not understand how this implies $\overline{A} \subset U^c$. My question arises since $A \subset \overline{A}$, how can we know that $\overline{A} \cap U = \emptyset$?
EDIT: I have been thinking about this a bit more and I think I have the solution:
Let $x \in \overline{A}\backslash A$ and assume towards contradiction that $x\not\in acc(A)$. Then there exists an open set $U$ such that $U \cap A = \emptyset$. Then $\overline{A} \cap U^c$ is a closed set containing $A$. Since $\overline{A}$ is the smallest set containing $A$, we have a contradiction unless $\overline{A}\subset U^c$ which requires $\overline{A}\cap U = \emptyset$.
Is this correct? Also it seems to be a lot to leave out in a proof that doesn't seem to be a sketch. Is there a more concise way of thinking about this?

I think your "EDIT" starts well, but doesn't finish the job. You never mentioned $x$ again, in particular the fact that it belongs to $U$. We are trying to prove the middle bit, that $\overline{A} \subset A \cup acc(A)$.
Let $x \in \overline{A}\backslash A$ and assume towards contradiction that $x\not\in acc(A)$. Then there exists an open set $U$, with $x$ in it, such that $U \cap A = \emptyset$. Then $\overline{A} \cap U^c$ is a closed set containing $A$. Since $\overline{A}$ is the smallest closed set containing $A$, we have  $\overline{A}\subset \overline{A} \cap U^c$ which implies $\overline{A}\subset  U^c$. Thus $x \not\in \overline{A}$, a contradiction.
