Dumb question: If $Y$ is a subspace of $X$, and $K \subseteq Y$ is compact in $Y$'s subspace topology, then I cannot necessarily conclude that $K$ is compact in $X$, right?
@user193319 One can conclude that: Let $\{U_i\}$ be an arbitrary cover of $K$ in $X$. Then $\{U_i \cap Y\}$ is an open cover of $K$ in $Y$ and by compactness one have finite subcover......
