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?

$(X,\tau ),$ 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 :  $\mathcal{U}(A):=\{U|(U\in \tau)(A\subseteq U)\}$
