Compactness and separation of topological space

Question

Let $X$ be a Hausdorff space, let $x \in X$ and $K \subseteq X$ be compact. Show that $x$ and $K$ can be separated by disjoint open sets i.e. show that there exist open $U,V \subseteq X$ with $U \cap V = \emptyset$ such that $x \in U$, $K \subseteq V$.

Answer 1

Let $x \in X$, and $K \subseteq X$ be compact. Let $y \in K$ be arbitrary, then since $X$ is Hausdorff there exist $V_y$ and $U_y$ open with $U_y \cap V_y = \emptyset$ in $X$ s.t. $x \in U_y$ and $y \in V_y$. Thus $\{V_y\}_{y \in Y}$ is an open cover of $K$ and since $K$ is compact, it has a finite subcover. $\{V_{y_i}\}_{i = 1}^n$.

So we have $K \subseteq \cup_{i = 1}^n V_{y_i}$ and $x \in \cap_{i = 1}^n U_{y_i}$. Both of the sets are open; in the latter case this is because the intersection is finite. Also, the two sets are disjoint which concludes the proof.