Prove Theorem 7.8: Let $X$ be a Hausdorff topological space and $A$ be compact in $X$. Then $A$ is closed in $X$. My book gave this proof, but I am trying to understand some of the reasoning behind each of the steps. After each step I  have numbers in brackets, denoting the places I don't understand the reasoning. I have provided some definitions, but am still trying to understand this proof.
Hausdorff: A topological space $X$ is Hausdorff if for every pair of distinct points $x$ and $y$ in $X$, there exists disjoint neighborhoods $U$ and $V$ of $x$ and $y$ respectfully.
7.1: Let $A$ be a subset of a topological space $X$ and let $O$ be a collection of subsets of $X$.
Cover: The collection $O$ is said to cover $A$ or to be a cover of $A$ if $A$ is contained in the union of the the sets in $O$
Open Cover: If $O$ covers $A$, and each set in $O$ is open, then we call $O$ and open cover of $A$.
Subcover: If $O$ covers $A$, and $O^\prime$ is a subcollection of $O$ that also covers $A$, then $O^\prime$ is called a subcover of $O$
Compact: A topological space $X$ is compact if every open cover of $X$ has a finite subcover.
Start of Proof:
Let $A$ be compact in the Hausdorff space X. [Given]
Want to show that $X - A$ is open [If the complement is open, the space is closed - Definition of closed]
Thus, let $x \in X-A$ be arbitrary [if $x$ is arbitrary true for any point]
Now, we need to show that there is an open set $U$ such that $x \in U \subset X-A$ [Theorem 1.4]
Since $X$ is Hausdorff, we know that for each $a \in A$, there exists disjoint open sets $U_a$ and $V_a$ such that $x \in U_a$ and $a \in V_a$. [Definition of Hausdorff]
Then $O = \{V_a\}_{a \in A}$ is an open cover of $A$. [1]
Because $A$ is compact [Given]
There is a finite subcover $\{V_{a_1},...,V_{a_n}\}$ of $O$. [If A is compact every open cover has a finite subcover by definition]
Let $V = \bigcup^n_{i=1} V_{a_i}$ [2]
and $U = \bigcap^n_i=1 U_{a_i}$ [3]
Then $U$ and $V$ are open sets such that $A \subset V$ and $x \in U$. [4]
Furthermore, since $U_{a_i}$ and $V_{a_i}$ are disjoint for each $i$, it follows that $U$ and $V$ are disjoint as well. [5]
Thus $U$ and $A$ are disjoint [6]
and therefore there exists an open set $U$ such that $x \in U \subset X-A$. [7]
Hence $X-A$ is open, implying that $A$ is closed.
 A: *

*The $V_a$ are open for each $a \in A$ (that's how it's chosen in the application of Hausdorffness) and each $a \in A$ is covered by "its own" $V_a$, so it's clear that $A \subseteq \bigcup \{V_a\mid  a \in A\}$ which is what it means to cover $A$.


*is the just reformulating (using some notations) that $\{V_a\mid a \in A\}$ has a finite subcover. So finitely many $V_a$ can be found (for diffrent $a \in A$ so that still $A$ is a subset of their union. The proof just gives a name to the $a$ whose $V_a$ are used in this finite subcover: there are $a_1, a_2, \ldots a_n$ (so $n\in \Bbb N$ is the number of sets in the finite subcover) so that $A \subseteq \bigcup_{i=1}^n V_{a_i} = \bigcup \{V_{a_i}\mid i=1,2,\ldots n\}$. The name $V$ is introduced for this finite union.


*For those sam  $a_1,a_2,\ldots,a _n$ we have corresponding $U_{a_i}$ which are their disjoint counterparts that all contain $x$ by construction. ($x$ is the fixed arbitary point outside $A$ that we're working with in this part of the proof), and so $U:= \bigcap_{i=1}^n U_{a_i}$ is a finite (this is essential) intersection of open sets that all contain $x$ and so $x \in U$ and $U$ is open.


*That $A \subseteq V$ is already part of the 2 and the definition of $V$ as the union of the finite subcover. $x \in U$ I noted in 3 already. No new info.


*$U \cap V = \emptyset$ is the whole point: If $p \in U \cap V$ then $p \in V_{a_j}$ for some $1 \le j \le n$ (definition of union) and also $p \in U_{a_j}$ for that some $j$ (as $p$ is in the intersection of $U_{a_i}$ ,so in all of them). Contradiction, as these sets $U_{a_j}$ and $V){a_j}$ are the $U_a$ and $V_a$ sets (for $a=a_j$) that were chosen disjointly by Hausdorffness! So no such $p \in U \cap V$ can exist.


*As $A$ is a subset of $V$, $U$ and $A$ are certainly disjoint too.


*So reformulated: $x \in U \subseteq X-A$ (two sets are disjoint iff one is a subset of the compelment of the other, a simple logical reformulation).
As each point of $X-A$ is an interior point of $X-A$, $X-A$ is open and is $A$ is closed.
A: Here's some guidance to each of your questions. Hopefully this helps!

*

*For all $a \in A$, there exist disjoint open sets $V_{a} \ni a$ and $U_{a} \ni x$. Since $A = \cup_{a \in A} \{a\} \subseteq \cup_{a \in A} V_{a}$ is a union of open sets that covers $A$, it is an open cover.

*Since $A$ is compact, all open covers (including this one) admit a finite subcover $\{V_{a_{j}}\}_{j=1}^{n}$ (that is, $A \subseteq \cup_{j=1}^{n} V_{a_{j}} = V$) for some $\{a_{j} \}_{j=1}^{n} \subseteq A$.

*Note that $U = \cap_{j=1}^{n} U_{a_{j}}$ is a finite intersection of open sets containing $\{x \}$ and so $x \in U \neq \emptyset$ is open.

*Since $V$ is an open cover of $A$, $A \subseteq V$. Since $x \in U_{a_{j}}$ for each $a_{j}$, it follows that $x \in U$.

*The set $U$ is disjoint from $V$ since $V_{a_{j}} \cap U \subseteq V_{a_{j}} \cap U_{a_{j}} = \emptyset$ for each $a_{j}$. Note that $V \cap U = \cup_{j=1}^{n} (V_{a_{j}} \cap U) = \emptyset$. This is just de Morgan.

*Finally $A \cap U \subseteq V \cap U = \emptyset$ and so $A, U$ are disjoint.

*For $x \in X \setminus A$, $\exists U \ni x$ such that $U \subseteq X \setminus A$ (or equivalently $U \cap A = \emptyset$).

*Hence, since $x$ is arbitrarily chosen from $X \setminus A$, it follows that $X \setminus A$ is open and $A$ is closed in $X$.

