To begin with, I know this question has been answered here multiple times. However, I want to gain some intuition about the proof and be lead to the right direction without being given the proof explicitly.
Let $A,B$ be disjoint compact subsets in a Hausdorff space.
I want to show $\exists \; U,V \; : \; A \subseteq U, \; B\subseteq V$
Where $U,V \subseteq X$ are disjoint and open
We know that subsets of Hausdorff spaces are also Hausdorff spaces. So every point in $A$ is contained in some open set that is disjoint from B (similarly for $B$). From here, we have open covers for $A$ and $B$, from which we can extract finitely many open sets to cover their respective sets by compactness.
So now we have disjoint finite covers for both sets, my goal now is to create a bigger set for $A$ and $B$ respectively. However I'm not quite sure if I'm heading in the right direction or if I've made a mistake anywhere.
Best Answer
Hint: It is helpful to first prove the following fact to be used in the proof: for any point $x \notin B$, there exist disjoint open sets $U,V$ with $x \in U$ and $B \subset V$.
Proof of the initial fact:
For each $b \in B$, there exist disjoint open sets $U_b,V_b$ such that $x \in U_b$ and $b \in V_b$. The set $\{V_b: b \in B\}$ is an open cover of $B$, so there exists a finite set $b_1,\dots,b_n$ such that $\{V_{b_1},\dots,V_{b_n}\}$ is a finite open cover of $B$.
Take $U = \bigcap_{j=1}^n U_{b_j}$ and $V = \bigcup_{j=1}^n V_{b_j}$. Verify that $U,V$ are disjoint open sets with $x \in U$ and $B \subset V$.