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> 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. 
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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. $\square$