Closed subsets of compact sets are compact (original proof) 
Baby Rudin Theorem 2.35: Suppose $F \subset K \subset X$ where $X$ a metric space. Suppose $F$ closed relative to $K$ and  $K$ compact. Then $F$ is compact. 

This is my original attempt to prove this theorem.
Here, Rudin's definitions are :


*

*$p$ is a limit point of a subset of a metric space iff every neighborhood(defined in terms of open balls) of $p$ contains a point in the subset which is not equal to $p$

*Subsets of metric spaces are closed iff it contains all its limit points.

*A point $p$ is an interior point of a subset of a metric space if there exists a neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within the subset. 

*Subsets of metric spaces are open iff all its points are interior points. 

*Subsets of metric spaces are compact iff every open cover has a finite subcover. 
Proof: Suppose $K$ compact and $F \subset K$ closed but not compact. Then there is a cover of $F$ which has no finite subcover say $S$. Let $G=\bigcup^{n}_{i=1}G_i$ be an open cover of $K$ which is finite. Define $M_i=G_i\cap F^c $ where $F^c$ complement of $F$. Then each $M_i$ is open since $G_i$ and $F^c$ are open. Now for any $x \in K$, either $x \in F$ or $x \in F^c$. If $x\in F$, then $x \in S_\alpha $ where $S_\alpha$ is an element of $S$. If $x \in F^c$ then $x \in G_i\cap F^c $ for some $i \in \mathbb{N}$ since $K \subset G$. So if $M=\bigcup^{n}_{i=1} M_i$, then $M \cup S$ is a cover of $K$. But then clearly, this cover has no finite subcover hence $K$ is not compact. $\square$

Refined proof:
Suppose $F$ not compact. Then there is a cover of $F$ which has no finite subcover say $S$. Define $M=\{X\}\setminus F$. Then $M \cup S$ is a cover of $K$. But then clearly, this cover has no finite subcover hence $K$ is not compact. $\square$
