I don't have the book handy, so I'll write down the proof in my own notation.
The proof is as follows: suppose $K$ is compact relative to $X$. We want to show it is compact relative to $Y$:
Take any open cover of $K$, $\{U_\alpha \mid \alpha \in A\}$. where all $U_\alpha$ are open in $Y$. As $Y$ has the subspace topology with respect to $X$, for every $\alpha$ we have $O_\alpha \subseteq X$ open such that $U_\alpha = O_\alpha \cap Y$. Every element $x \in K$ is in some $U_\alpha$, and so in the corresponding $O_\alpha$ as well, so $\{O_\alpha \mid \alpha \in A \}$ is also a cover of $K$ by open sets from $X$. Now the assumption that $K$ is compact relative to $X$ can be used, so there is a finite subset $B \subseteq A$, such that $K$ is covered by $\{O_\alpha \mid \alpha \in B\}$. Now if $x \in K$, we know that there is some $\beta \in B$ such that $x \in O_\beta$. As $x \in K \subseteq Y$ we also know that $x \in O_\beta \cap Y = U_\beta$. This shows that $\{U_\alpha \mid \beta \in B\}$ is the required finite subcover of our original cover.
Now suppose that $K$ is compact relative to $Y$, and we want to show $K$ is compact relative to $X$ as well. So take an open cover $\{O_\alpha \mid \alpha \in A\}$ again, where all $O_\alpha \subseteq X$ are open in $X$. Now define $U_\alpha = Y \cap O_\alpha$, then by the definition of the subspace topology on $Y$ we know that all $U_\alpha$ are open in $Y$. As before, if $x \in K$, $x \in O_\alpha$ for some $O_\alpha$ and $x \in Y$, as $K \subseteq Y$, so $x \in O_\alpha \cap Y = U_\alpha$ for that $\alpha$, which shows that $\{U_\alpha \mid \alpha \in A\}$ is a cover of $K$ by open subsets from $Y$. So we apply the assumption and get a finite subset $B \subseteq A$ such that $\{U_\beta \mid \beta \in B\}$ is a finite subcover. If $x \in K$, than for some $\beta \in B$, $x \in U_\beta = O_\beta \cap Y$, so $x \in O_\beta$ for that $\beta$, which shows that $\{O_\beta \mid \beta \in B\}$ is the required finite subcover for our original cover.
