Let $(X, d)$ be a metric space, $Y$ ⊂ $X$ and consider the metric space $(Y, d)$.
Show that every open set $U$ in $Y$ has the form $U$ = $V$ ∩ $Y$ for an open set $V$ ⊂ $X$.
Show that $Y$ is compact in $(Y, d)$ if and only if every collection of open sets { Vk } k ∈ K in $X$ that covers $Y$ has a finite subcover.
I'm beyond confused on how to go about this problem. I mostly understand what it is that I have to show, but have no idea how to show any of it.
Hint: 1) $U\subset Y$ is open implies that $U$ is a union of open balls in $Y$. What happens if you take the same open balls, but see them as balls in $X$?
2) If $Y$ is compact, then $\{V_k \cap Y\}_K$ is an open cover of $Y$ by opens in $Y$, so by compactness... If every cover of $Y$ by opens of $X$ admits a finite subcover, then use part 1) to show that $Y$ is compact.