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1.6.4. Lemma. If $C$ is a compact subset of a Hausdorff topological space $(X, \tau)$, there are for each $x \not \in C$ disjoint open subsets $A$ and $B$ of $X$, such that $C \subset A$ and $x \in B$.

Proof: Since $X$ is a Hausdorff space, there are for each $y$ in $C$ disjoint open sets $A(y)$ and $B(y)$, such that $y \in A(y)$ and $x\in B(y)$. The family $\{A(y) \ \vert \ y \in C \}$ is an open covering of $C$ and has therefore a finite sub covering $A(y_1), A(y_2), ..., A(y_n)$. Set $A = \bigcup A(y_k)$ and $B = \bigcap B(y_k)$.

I want to check that $A$ and $B$ are in fact disjoint.

Therefore, suppose that $A \cap B \neq \emptyset$. Then $$\left( \bigcup_k A(y_k) \right) \cap \left( \bigcap_k B(y_k) \right) \neq \emptyset.$$ I'm confused by the presence of two intersections, and haven't been able to proceed.

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    My professor once said: when you find hard notation, call $T$ what gives you a problem. Try defining $T := \bigcap_k B(y_k)$; note that $T \in \tau$ – 2017-01-06

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The author simply substituted for $A$ and $B$ the definitions of these two sets. However, you don’t actually have to write that out; you can argue directly as follows.

If $A\cap B\ne\varnothing$, then there is some $x\in A\cap B$. Then $x\in A=\bigcup_kA(y_k)$, so there is a $k_0$ such that $x\in A(y_{k_0})$. And $x\in B=\bigcap_kB(y_k)$, so $x\in B(y_k)$ for every $k$. In particular, $x\in B(y_{k_0})$. But then $x\in A(y_{k_0})\cap B(y_{k_0})=\varnothing$, which is clearly impossible. This contradiction shows that $A\cap B$ must be empty.