$U,V$ are disjoint open sets in $X$ and $P\subset \overline{U}\cap \overline{V}$, then $P\cap(U\cup V)=\varnothing$ Is this statement always true?
Let $U,V$ be disjoint open sets in $X$ and $P\subset \overline{U}\cap \overline{V}$, then $P\cap(U\cup V)=\varnothing$.
 A: If $x\in P$ then in every open neighborhood of $x$ there are elements from $U$ and $V$. If $x\in U$ then it had an open neighborhood which meets $V$ as well, but that is impossible (can you see why?), similarly if $x\in V$.
Therefore if $x\in P$ then $x\notin U\cup V$, and so the intersection is indeed empty.
A: I think you can choose $P$ maximal so $P=\overline{U}\cap \overline{V}$. Expanding 
leads to 
$$ (\partial U \cap V)\cup (\partial V \cup U) \cup (\partial V \cap \partial U)\cap (U \cup V)=$$ 
Since $U,V$ are open and disjunct the first two terms don't matter (maybe incorrect) so we get
$$ (\partial V \cup \partial U) \cap (U \cup V) $$. This is the same as 
$$((\partial U \cup \partial V)\cap U )\cup ((\partial U \cup \partial V) \cap V)$$, and again we have 
$$ (\partial U \cap V) \cup (\partial V \cap U)$$.
So yeah it its $\varnothing$
A: If by $\bar{A}$ you mean the closure of $A$, then this is obviously not true, as long as $P\neq\emptyset$.
If else you mean the complement of $A$, which is usually denoted by $A^c$, it is always true. You have $(P\cap(U\cup V))^c = P^c\cup(U^c\cap V^c) = X$ by assumption, and by noting that $(A^c)^c=A$, you get your statement.
