I was looking over Theorem 2.33 from Rudin's Principles of Mathematical Analysis, and I was wondering why the sets $G_{\alpha}$ are open relative to $X$? I thought that by definition, the sets $G_{\alpha}$ should be open in $X$. I've provided Theorem 2.30 and the proof of Theorem 2.33 below.
As reference: Theorem 2.30 Suppose $Y \subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.
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 $Y$.
Proof Suppose $K$ is compact relative to $X$, and let $\{V_{\alpha}\}$ be a collection of sets, open relative to $Y$, such that $$K \subset \bigcup_{\alpha} V_{\alpha} \tag{22}$$ By theorem 2.30, there are sets $G_{\alpha}$, open relative to $X$, such that $V_{\alpha} = Y \cap G_{\alpha}$, for all $\alpha$; and since $K$ is compact relative to $X$, we have $$K \subset G_{\alpha_{1}} \cup \cdots \cup G_{\alpha_{n}}\tag{23}$$ for some choice of finitely many indices $\alpha_{1}, \cdots ,\alpha_{n}$. Since $K \subset Y$, $(22)$ implies $$K \subset V_{\alpha_{1}} \cup \cdots \cup V_{a_{n}}$$ This proves that $K$ is compact relative to $Y$.