I'm trying to understand the following claim in a proof that $\text{Bd}(A) \cap \text{Int}(A) = \emptyset$, as well as a claim in the proof of $\overline{A} = \text{Int}(A) \cup \text{Bd}(A)$. I will bold the misunderstandings below.
First misunderstanding: If $x \in \text{Int}(A)$, then $x \in U$ for some open set $U \subset A$. Then, $U$ is a neighbourhood of $x$ disjoint from $X-A$, so $x$ is not in the closure of $X-A$.
Second misunderstanding: It is obvious that $\text{Int}(A) \cup \text{Bd}(A) \subset \overline{A}$. (Why is this obvious?)
If anyone can help clarify, that would be greatly appreciated.
(1). We have $Int (A)\subset A\subset \overline A.$ And we have $Bd(A)=\overline A\cap \overline {X\backslash A}\subset \overline A.$
So the union $Int(A)\cup Bd(A)$ is also a subset of $\overline A.$
(2a). The closure $\overline Y$ of a set $Y$ is defined as the common intersection of all closed sets that have $Y$ as a subset. So the complement, in $X,$ of $\overline Y$, is the union of all open sets that are disjoint from $Y.$
(2b). So any open subset of $X$ that is disjoint from $Y=X-A $ is also disjoint from $\overline Y=\overline {X-A}.$ In your Q, we have open $U$ disjoint from $X-A,$ so $U$ is disjoint from $\overline {X-A}.$ And $p\in U.$
Remark. From the last sentence of (2a) we have $X-\overline Y=Int (X-Y).$ In particular, if $Y=X-A, $ then $X-\overline {X-A}=Int (X-(X-A))=Int(A)$.
This tells us that $\overline A- Int(A)=$ $\overline A - (X-\overline {X-A})=$ $=\overline A \cap \overline {X-A}$ $=Bd(A).$ That is worth repeating: $$\overline A-Int(A)=Bd(A).$$
So, since $\overline A\supset A\supset Int(A)$, we have $\overline A=(\overline A -Int A)\cup Int(A)=Bd(A)\cup Int (A).$
This also tells us that we don't need (1) to prove that $\overline A=Int(A)\cup Bd(A).$