Prove that if $\ K\subset\mathbb R ^n$ is compact and $\ F\subset K$ is closed then $\ F$ is compact. My attempt:
Let$\ U$ be an open cover of$\ F$.
Every open subset of$\ U$ has the form $\ U\cap F$.
Let$\ V=(U\subset K : U$ is open and $\ \exists U' \in U $ such that $\ U \cap F = U'$ ) then$\ V$ is an open cover of$\ F$.
I'm stuck at this point, how can I go on with the proof? help would be greatly appreciated. 
 A: Take an open cover $O_\alpha$ of $F \subset K \subset \mathbb{R}^n$. We see that because $F$ is closed, $\mathbb{R}^n \setminus F$ is open, and hence:
$$
(\mathbb{R}^{n}\setminus F) \cup \bigcup _{\alpha}O_\alpha
$$
Forms an open cover of $K$ (it covers all of $\mathbb{R}^{n}$ actually). We can extract a finite subcover of $K$ from this cover (because $K$ is compact), consisting of $O_1, \cdots ,O_n$ and then possibly $\mathbb{R}^n \setminus F$. Note then that the $O_1, \cdots , O_n$ have to cover $F$, because $(\mathbb{R}^{n} \setminus F )\cap F = \emptyset$. We have shown then that for the open cover $O_\alpha$, we can extract a finite subcover, and the result follows.
Note that this statement is true in any topological space (any set where the terms "open", "closed" are defined) and hence is true outside of $\mathbb{R}^n$
A: In $\mathbb R^n$, compactness is equivalent to be closed and bounded. So if $K$ is compact, $K$ is closed and bounded. $F \subset K$ must be bounded. It follows $F$ is compact.
