Proving that a compact subset of a Hausdorff space is closed I am having trouble understanding the answers here.
I am trying to prove that a compact subset of a Hausdorff space is closed.
Following the proof is difficult, perhaps because Brian reused letters for different things(although I get they are arbitrary, I can't follow it.) The second answer uses nets and filters, which I don't know of, so would like to avoid for now.

Now trying to follow Brians proof, I get something like the following:
Let $S$ be our Hausdorff space, and $C\subset S$ be a compact subspace of $S$. (easy to remember, think $S$ for space, $C$ for compact)
To show that this subspace is closed, we can show the complement of the subspace is open. I.e. $S/C$ is open. Now following the proof of Brian, we take some element $x\in S/C$ and since $S$ is Hausdorff(which means $C$ is Hausdorff), we have for each $y\in C$ that there are disjoint open sets $U_y,V_y$ such that $x\in U_y$ and $y\in V_y$ now here is a little confusing, either he has reused $x$, or for some reason $x\in C$, which shouldn't be possible, since $x\in S/C$. Let's assume this is reusing the letter. Then the collection of all of these disjoint open sets, lets call it $\{\alpha_i,i\in C\}$ is an open cover of $C$, and since $C$ is compact, it has a finite subcover, $\{\alpha_i|u\in F\}$ where $F$ is a finite subset of $C$. Let $$U=\bigcap_{x\in F}V_y=\emptyset$$
Right? So then $U$ is open, since all empty sets are open... Now the proof doesn't even seem to follow.

I have no idea what's going on!
 A: Here’s a sketch that may help a little.

The large black oval represents the compact set $C$, $x$ is the point outside $C$, and $y$ and $z$ are two points of $C$ that I’ll use to illustrate the choice of open nbhds. The space is Hausdorff, so $x$ and $y$ have disjoint open nbhds; in the picture they’re the smaller black ovals, $V_y$ being the nbhd of $y$, and $U_y$ the corresponding nbhd of $x$. Similarly, $x$ and $z$ have disjoint open nbhds; I’ve drawn them in blue, $V_z$ being the nbhd of $z$, and $U_z$ being the corresponding nbhd of $x$.
The idea is to do this for each point of $C$. Thus, corresponding to each point $p\in C$ we get two open sets, which I’ve called $U_p$ and $V_p$: $V_p$ is an open nbhd of $p$, $U_p$ is an open nbhd of $x$, and the sets $U_p$ and $V_p$ are disjoint. I can do this because the space is Hausdorff.
The sets $V_p$ with $p\in C$ form an open cover $C$: each of them is open, and for any $p\in C$ I know that $p\in V_p$, so they cover $C$. $C$ is compact, so only finitely many of them are needed in order to cover $C$. This means that there is a finite subset $\{p_1,\ldots,p_n\}$ of $C$ such that $$C\subseteq V_{p_1}\cup\ldots\cup V_{p_n}\;.$$
Now I look at the corresponding open nbhds of $x$, $U_{p_1},\ldots,U_{p_n}$. Specifically, I let
$$U=U_{p_1}\cap\ldots\cap U_{p_n}\;;$$
since $U$ is an intersection of only finitely many open sets, $U$ is also open. I claim that $U\cap C=\varnothing$. Suppose, on the contrary, that there is a point $y\in U\cap C$. We saw above that 
$$C\subseteq V_{p_1}\cup\ldots\cup V_{p_n}\;,$$
so there is a $k\in\{1,\ldots,n\}$ such that $y\in V_{p_k}$. But $U\subseteq U_{p_k}$, and $y\in U$, so $y\in U_{p_k}$, which is impossible: we chose $U_{p_k}$ and $V_{p_k}$ to be disjoint (like $U_y$ and $V_y$ in the sketch, or $U_z$ and $V_z$). Thus, no such point $y$ exists, and $U$ is therefore an open nbhd of $x$ disjoint from $C$.
Since we can do this for any point $x$ not in $C$, $C$ must be closed.
A: Alternative write up to Brian's(as interpreted from Brian's write up):

Let $A$ be a compact subset of the Hausdorff space $X$. Take $x\in A$ and $y\in X/A$. Then $x\ne y$ and by the definition of Hausdorff spaces, there are two disjoint open sets $G_{x,y}$ and $H_{x,y}$ of $X$ such that $x\in G_{x,y}$ and $y\in H_{x,y}$ with $G_{x,y}\cap H_{x,y}=\emptyset$
Since $\{G_{x,y}|x\in A\}$ for each $y\in X/A$ is an open cover of $A$, i.e. we have $A\subset \cup_{x\in A} G_{x,y}$. Since we have compactness for $A$, there is a finite subcover of $A$ consisting of the elements $x_1,\cdots,x_n$ of $A$, such that $$A\subset \cup_{i=1}^n G_{x_i,y}$$
Also for each $i=1,\cdots,n$ there is an open set $H_{x_i,y}$ such that $y\in H_{x_i,y}$. Let $H_y = \cap_{i=1}^n H_{x_i,y}$ and note that $y\in H_y$ and $H_y$ is an open set of $X$. Furthermore, $H_y\cap(\cup_{i=1}^n G_{x_i,y})=\emptyset$
Thus $H_y\subset X/A$ and therefore $X/A = \cup_{y\in {X/A}} H_y$ is an open set and so $A$ is closed.
