Reviewing for qual:
Let $X$ be a Hausdorff space, $K$ a nonempty compact subset of $X$, and $x \in X\backslash K$. Prove that there exist disjoint, open subsets $U$ and $V$ such that $K \subset V$ and $x \in U$.
The solution we have is as follows:
For any $y \in K$, there are open neighborhoods $U_y$ with $x \in U_y$ and $y \in V_y$ with $U_y \cap V_y = \emptyset$. Since $K$ is compact, the cover {$V_y$}$_{y \in K}$ has a finite subcover: $V_{y_1}, \cdots, V_{y_n}$. Let $U = \bigcap_{j=1}^n U_{y_j}$ and $V = \bigcup_{j=1}^n V_{y_j}$. Then $x \in U$, $K \subset V$, and $U \cap V = \emptyset$
So I have two questions about the solution:
$1$) Why is $U \cap V = \emptyset$?
$2$) Why do we need to let $U$ be an intersection of all such $U_{y_j}$? I'm trying to understand why we need to let $U$ be an intersection and $V$ be a union (I assume the union is ok because we want $K$ to be covered by the union of finitely many open sets, so it's more about why $U$ needs to be an intersection).
Suppose $x\in U\cap V$. Then $x\in V=\bigcup_{j=1}^nV_{y_j}$, so $x\in V_{y_{i}}$ for some $i$. But we also have $x\in U=\bigcap_{j=1}^nU_{y_j}\subseteq U_{y_i}$, so that $x\in U_{y_i}\cap V_{y_i}$ and this contradicts that $U_{y_i}\cap V_{y_i}=\varnothing$.
We let $U$ be the intersection of the $U_{y_j}$ in order to achieve that $U\cap V=\varnothing$. That would not have to be the case if e.g. $U$ was taken to be the union of the $U_{y_j}$.