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?

$A \cap U = \emptyset$ is equivalent to $A \subseteq U^\complement$. The set $U^\complement$ is closed, as $U$ is open. It follows that $\overline{A} \subseteq \overline{U^\complement} =  U^\complement$, as claimed.
