The theorem is that: Suppose $K \subset Y \subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$. Rudin begins the proof by stating that:
Suppose $K$ is a compact relative to $X$, and let $\{V_{\alpha}\}$ be a collection of sets, open relative to $Y$, such that $K \subset \cup_{\alpha}V_{\alpha}$.
The opening statement makes me feel uneasy, as we are directly taking an open cover for $K$, whose sets $V_{\alpha}$ are also open relative to $Y$. Namely, when we assume that $K$ is open relative to $X$, then by definition for any open cover for $K$ in $X$ there exists a finite subcover. But why are the sets of the open cover (note: not the subcover) necessarily open in $Y$? I say necessarily, as I have understood the process for showing the theorem to be true that i.) assume that $K$ is compact in $X$, take an arbitrary open cover for $K$ in $X$ so it has a finite subcover. Then show that any open cover for $K$ in $Y$ also has a finite subcover, so that $K$ is compact in $Y$. I don't quite understand why there should be a linkage between the openness of the cover of $K$ in $X$ and openness of the (same) cover relative to $Y$.
Best Answer
The argument is a bit different than the one you've outlined at the end of your question. It goes like this:
Suppose $K$ is compact relative to $X$. We will show that $K$ is compact relative to $Y$. It is enough to show that for any open (relative to $Y$) cover of $K$, there is a finite subcover. Let $\{V_\alpha\}$ be a collection of sets, open relative to Y, such that $K \subset \cup_{\alpha}V_{\alpha}$. We will show that $\{V_\alpha\}$ has a finite subcover.