Prove that $\{(x,y) \in \mathbb R^2 | y = x^2 \}$ is not compact I know I need to choose an open cover and then show it has no finite subcover. 
If I use $((-n,-n^2),(n,n^2)) \forall n \in \mathbb N$ does this work?
 A: Hint : prove that the given set is not bounded.
A: What do you mean by $(\langle -n,-n^2\rangle,\langle n,n^2\rangle)$? If you’re talking about a cover by open sets in $\Bbb R^2$, you’re using interval notation for what are supposed to be open sets in $\Bbb R^2$, which doesn’t have a linear structure that would allow interval notation to be meaningful. If you’re talking about sets that are open in the relative topology of $S=\{\langle x,y\rangle:y=x^2\}$, then your interval notation could be made meaningful, but you’ve not done so. (In any case I suspect that you mean the lefthand end to be $\langle -n,(-n)^2\rangle=\langle -n,n^2\rangle$, not $\langle -n,-n^2\rangle$, since the latter point isn’t even in $S$.)
You can make something very similar to your idea work. For $n\in\Bbb Z^+$ let 
$$U_n=\{\langle x,y\rangle\in\Bbb R^2:-n<x<n\}=(-n,n)\times\Bbb R\;;$$
this is open in $\Bbb R^2$, and clearly $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ covers $S$. (Indeed, it covers all of $\Bbb R^2$.) But if $\{U_{n_1},\ldots,U_{n_k}\}$ is a finite subset of $\mathscr{U}$ with $n_1<\ldots<n_k$, then $U_{n_1}\cup\ldots\cup U_{n_k}=U_{n_k}$, and $\langle n_k,n_k^2$ is a point of $S$ not in $U_{n_1}\cup\ldots\cup U_{n_k}$. Thus, no finite subset of $\mathscr{U}$ covers $S$, and $S$ is therefore not compact.
Of course if you’ve learned the theorem that a set in $\Bbb R^n$ is compact if and only if it’s closed and bounded, you can simply point out that $S$ isn’t bounded and therefore cannot be compact.
A: Let A >0 arbitrary . Consider $n \in N$ such that $\sqrt{(n)^2 + (n^2 - 0 )^2} > A$.
We have $(n,n^2)$ an element of your set  and the norm of this vector greater than A. Then your set is not bounded. And then is not compact .
