Suppose $K\subset Y \subset X$. Then, $K$ is compact relative to X $\textit{iff}$ $K $ compact relative to Y.
Solution: The proof in the Rudin goes like this:
In this proof, I understand that there exist some $G_{\alpha}$ for each $V_{\alpha}$ but how to reach to the fact that $K$ is subset of finite collection of these $G_{\alpha}$'s and not other open sets in $X$ (or RHS of (22) is the open cover of $K$). Eq. (22) is what I don't understand.
Problem rephrase (Edit): For a set to be compact, we have to show that every open open cover of $K$ contains finite subcover. i.e. If ${G_{\alpha}}$ is an open cover of $K$ , then there are finitely many indices $\alpha_1,\cdots,\alpha_n$ such that $K \subset G_{\alpha_1} \cup \ldots \cup G_{\alpha_n}$. Now, we have proven that there are some open sets in $X$ correspoding to each $V_{\alpha}$. How do we make sure that these $G_{\alpha}$ form an open cover?
[This the from rudin's book section-2.33]