Baby Rudin Theorem 2.33 Theorem 2.33
Suppose $K \subset Y \subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.
I understand the first part of the proof: if $K$ is compact relative to $X$ then $K$ is compact relative to $Y$.
However, the proof of the converse statement confuses me.
Proof given:
Conversely, suppose $K$ is compact relative to $Y$, let ${G_\alpha}$ be a collection of open subsets of $X$ which covers $K$, and put $V_\alpha = Y \cap G_\alpha$. Then $K \subset V_{\alpha_{1}} \cup ... \cup V_{\alpha_{n}}$ will hold for some choice of $\alpha_1, ..., \alpha_n$; and since $V_\alpha \subset G_\alpha$, $K \subset V_{\alpha_{1}} \cup ... \cup V_{\alpha_{n}}$ implies $K \subset G_{\alpha_{1}} \cup ... \cup G_{\alpha_{n}}$.
My question:
What is the purpose of introducing $V_\alpha$? Is it to show that $V_\alpha \in Y$ so that we can prove that $K$ is compact relative to $Y$?
 A: In this proof, you are not trying to prove that $K$ is compact relative to $Y$. 
Instead, you are assuming that $K$ is compact relative to $Y$, and you want to use that assumption to prove that $K$ is compact relative to $X$.
You've started off correctly: to prove $K$ is compact relative to $X$, you start by choosing $G_\alpha$ to be a collection of open subsets of $X$ which covers $K$.
Now comes the question: How do you identify a finite number of the $G_\alpha$ which still covers $K$? 
There's really only one thing to try, namely to somehow use the knowledge that $K$ is compact relative to $Y$.
Question: How shall you use that knowledge?
Answer: Take the intersection of each $G_\alpha$ with $Y$. 
What you get is a collection of sets $V_\alpha = Y \cap G_\alpha$. 
And now put together what you know about the sets $V_\alpha$: (1) You know that the $V_\alpha$ sets cover $Y$; and (2) You know that the $V_\alpha$ sets are open relative to $Y$. 
Applying the knowledge that $K$ is compact relative to $Y$, you may conclude that there exist $\alpha_1,\ldots,\alpha_n$ such that $K \subset V_{\alpha_1} \cup\cdots\cup V_{\alpha_n}$, and the proof is completed as you wrote it.
A: A bit more formal:
0) $K \subset Y \subset X$;
Show that: If $K$ is compact in $Y$ then $K$ is compact in $X$.
1) Consider an open  cover of $K$ in $X$, i.e.
$K \subset \bigcup G_{\alpha}$, $G_{\alpha} \subset X$, open in $X$.
2) $V_{\alpha}:= G_{\alpha} \cap Y$ is open in $Y$.
3) $K \subset \bigcup V_{\alpha}$ (why?).
4)Since $K$ is compact in $Y$ there is a finite subcover
$K \subset  V_{\alpha_1}\cup V_{\alpha_2} \cup...\cup V_{\alpha_n}$.
5) $V_{\alpha_i} \subset G_{\alpha_i}$ , $i=1,2,..,n$,
$K \subset \bigcup G_{\alpha_i}$, $i=1,2,..n$, a finite subcover of $\bigcup G_{\alpha}$, and we are done.
Note:
3') $K= K \cap Y \subset (\bigcup G _{\alpha}) \cap Y$.
$K \subset \bigcup (G_{\alpha} \cap Y)$(Distributive law).
