Compact Hausdorff spaces are normal I want to show that compact Hausdroff spaces are normal.
To be honest, I have just learned the definition of normal, and it is a past exam question, so I want to learn how to prove this:

I believe from reading the definition, being a normal space means for every two disjoint closed sets of $X$ we have two disjoint open sets of $X$.

So as a Hausdorff space, we know that $\forall x_1,x_2\in X,\exists B_1,B_2\in {\Large{\tau}}_X|x_1\in B_1, x_2\in B_2$ and $B_1\cap B_2=\emptyset$
Now compactness on this space, means we also have for all open covers of $X$ we have a finite subcover of $X$.

Now if we take all of these disjoint neighborhoods given by the Hausdorff condition, we have a cover of all elements, I am not sure how to think of this in terms of openness, closedness.

How does one prove this?
 A: We'll do this in two parts.
$\textbf{Lemma.}$ Let $T$ be Hausdorff. Suppose $x\in T $ and $Y \subset T$ be compact and let $x \notin Y$. Then there are open sets $U$, $V$ separating x and Y, i.e., $x\in U$ and $Y\subset V$ such that $U\cap V=\varnothing$.
Proof of the lemma:
Since the space is Hausdorff, for every $y\in Y$ there is a neighborhood $V_y$ of $y$ and a neighborhood $U_y$ of $x$ such that $U_y\cap V_y=\varnothing$. We have that $\bigcup_{y\in Y} V_y$ is an open cover of Y and since compact, there exists a finite subcover $ V_{y_1},V_{y_2},\dots,V_{y_n} $ for Y. Now let
\begin{align*}
U= \bigcap_{j=1}^n U_{y_j}, \qquad V= \bigcup_{j=1}^n V_{y_j}.
\end{align*}
Then $x\in U$ and $Y\subset V$ where $U\cap V =\varnothing$. To show that they really are disjoint assume there is $z\in U\cap V$. Then there exists $z\in V_{y_j}$ for some $j$ and $z\in U_{y_j}$ for all $j$. But $U_{y_j}\cap V_{y_j} =\varnothing$.
$\textbf{Main proof:}$
Let $T$ be a compact Hausdorff topological space and let $X$ and $Y$ be disjoint closed sets in $T$. By the lemma, for any $y\in Y$ there exists a neighborhood $U_y\ni y$ and an open set $O_y$ containing $X$ such that $U_y\cap O_y =\varnothing$.
Since $Y$ is a closed subset of a compact set it is itself compact, and therefore the cover $\left\lbrace U_y\right\rbrace _{y\in Y}$ of $Y$ has a finite subcover $U_{y_1},U_{y_2},\dots , U_{y_n}$.
The open sets
\begin{align*}
O^1=\bigcap_{j=1}^n O_{y_j} \supset X,  \qquad O^2=\bigcup_{j=1}^n U_{y_j} \supset Y,
\end{align*}
are then disjoint open sets containing $X$ and $Y$, proving that every compact Hausdorff space is normal.
A: I will use the following proposition:
In a Hausdorff space, it is possible to separate a compact subset K and a point p not in K by disjoint open sets.
Assume $S_1,$ and $S_2$ are closed so they are compact. Apply proposition above to $S_1$ and $x \in S_2$ to get open sets:
$V_x,U_x$ such that
$$S_1 \subset V_x \ \ \ and \ \ \ x \in U_x$$
Therefore, $$S_1 \subset \bigcup_{x \in S_2} U_x$$ By compactness, we have $S_1 \subset \bigcup_{i = 1}^n U_{x_i} = U$. Set $V = \bigcap_{i = 1}^n V_{x_i}$, then it follows:
$$V \cap U = \emptyset$$
$$S_1 \subset V \ \ \ and \ \ \ S_2 \subset U.$$
A: The standard approach is to first show that such a topological space is regular. The argument to show that the space is normal is an easy adjustment of this argument, so I will present this proof instead.
Let $X$ be a compact Hausdorff space and let $F\subset X$ be a closed set. Let $y\in X\setminus F$. We need to construct open sets $U$ and $V$ with $F\subseteq U$ and $y\in V$ such that $U\cap V=\emptyset$. Since $X$ is Hausdorff, for each $x\in F$ there exists neighbourhoods $U_x$ of $x$ and $V_x$ of $y$ such that $x\in U_x$ and $y\in V_x$ and $U_x\cap V_x=\emptyset$. Let $\mathcal F$ be the collection
$$
\mathcal F=\{U_x:x\in F\}.
$$
Then $\mathcal F$ is an open cover of $F$ and $F$ is compact (why?) so there exists finitely many points $x_1,\dotsc,x_n\in F$ such that $F\subseteq \bigcup_{j=1}^n U_{x_j}$. Set $U=\bigcup_{j=1}^n U_{x_j}$ and let $V=\bigcap_{j=1}^n V_{x_j}$. Now $U$ and $V$ are open sets and $F\subseteq U$ and $y\in V$. Moreover, since $U_x\cap V_x=\emptyset$ for each $x\in F$, it follows that $U\cap V=\emptyset$. This shows that $X$ is regular.
A: Outline: Start with proving regularity i.e. for $x\in X$ and a closed subset $A\subset X$ not containing $x$, there are disjoint open subsets $U,V\subset X$, such that $x\in U$ and $A\subset V$. For this, note that $A$ is compact, as a closed subset of a compact space. Since $X$ is Hausdorff, for every $a\in A$ there are disjoint open subset $U_a,V_a$, with $x\in U_a$ and $a\in V_a$. The $V_a$'s cover $A$, and a simple argument shows regularity.
Then, using regularity, a process very similar to the one in the above paragraph shows normality.
A: $(X,\tau ),$ compact space; $\left( X,\tau \right) ,$ Hausdorff; $A\in \mathcal{C}\left( X,\tau \right) ;$ $B\in \mathcal{C}\left( X,\tau \right) ,$ $A\cap B=\emptyset,$ $x\in A$ and $y\in B.$
$\left.\begin{array}{r}\left( x\in A\right) \left( y\in B\right)  \\ \\ A\cap B=\emptyset \end{array}\right\}\Rightarrow \!\!\!\!\!\begin{array}{c}\mbox{} \\ \mbox{} \\ \left.\begin{array}{c} x\neq y \\ \mbox{} \\ {\left( X,\tau \right) ,\text{ }T_{2}}\end{array}\right\}\Rightarrow \!\!\!\!\!\end{array}$
$\left.\begin{array}{r}\Rightarrow \left( \exists U_{x}\in \mathcal{U}\left( x\right) \right) \left( \exists V_{y}\in \mathcal{U}\left( y\right) \right) \left( U_{x}\cap V_{y}=\emptyset \right) \\ \\ \left( \mathcal{A}:=\left\{ U_{x}|x\in A\right\} \right) \left( \mathcal{B}:=\left\{ V_{y}|y\in B\right\} \right) \end{array}\right\} \Rightarrow$
$\left.\begin{array}{r}\Rightarrow \left( \mathcal{A}\subseteq \tau \right) \left( A\subseteq \cup \mathcal{A}\right)\left( \mathcal{B}\subseteq \tau \right) \left( B\subseteq \cup \mathcal{B}\right)  \\ \\ \left( (X,\tau )\text{ is compact space}\right) \left( A,B\in \mathcal{C}\left( X,\tau \right) \right) \Rightarrow \left( A\text{ is }\tau \text{-compact}\right)\left( B\text{ is }\tau \text{-compact}\right)\end{array}\right\} \Rightarrow$
$\left.\begin{array}{r}\Rightarrow \left( \exists \mathcal{A}^{\ast }\subseteq \mathcal{A}\right) \left( \left\vert \mathcal{A}^{\ast} \right\vert <\aleph _{0}\right) \left( A\subseteq \cup \mathcal{A}^{\ast}\right) \left( \exists \mathcal{B}^{\ast }\subseteq \mathcal{B}\right) \left( \left\vert \mathcal{B}^{\ast }\right\vert <\aleph _{0}\right) \left( B\subseteq \cup \mathcal{B}^{\ast}\right)\\ \\ \left( U:=\cup \mathcal{A}^{\ast }\right) (V:=\cup \mathcal{B}^{\ast })\end{array}\right\} \Rightarrow$ 
$\left.\begin{array}{c}\Rightarrow \left( U\in \mathcal{U}(A)\right)\left( V\in \mathcal{U}(B)\right)\left( U\cap V=\emptyset \right).\end{array}\right. $
NOTE :  $\mathcal{U}(A):=\{U|(U\in \tau)(A\subseteq U)\}$
