Show that the extended complex plane is compact. Let $G$ be a subset of the extended complex plane (also known as the Riemann sphere see e.g. here), which we will denote as $(X, d)$ henceforth. Let $V \subset S$ denote the set of points which correspond to points $G \subset X$. Show that $V$ is open relative to $S$ iff $G$ is open relative to $X$. Now, conclude that the extended complex plane is compact.
I normally include some of my own thoughts on problems when posting. However, I have made very little progress aside from showing that the unit sphere (as a subset of $\mathbb{R}^3$), is itself compact.
Note: I have indeed looked at this previous Math.SE question and have not reached an understanding. I believe the answer is given via some topological definitions which I have not yet encountered. The context of this question is in a real analysis course corresponding to roughly the first half of Rudin.
 A: Sorry, my comments above are certainly incorrect. $V\neq S\cap G$ because $S$ is the representation of the points in $G$, not the points themselves. 
However, suppose that $S$ relative to $Y$ and $G$ closed relative to $X$. Then $G$ contains all its limit points. Let $p\in G$ be such a point. Then the representation of $p$ on the Riemann sphere is not in $V$. But this is a contradiction since by definition $V$ corresponds to all the points in $G$. Conversely, suppose $G$ open and $S$ closed. If $q$ is a limit point of $S$, then its representative on the plane is not in $G$, which is again a contradiction. 
Now, let $\mathcal{R}=\{\vec{x}\in\mathbb{R}^3:|\vec{x}|=1\}$ be the Riemann Sphere which is compact under the usual metric (I'll leave this proof up to you). So $\mathcal{R}$ has finite subcover, say $\bigcup_\alpha^n V_\alpha$. But by definition each $V_i$ corresponds to the set $G_i$ on the plane, where $G_i$ is open, so it follows that $\bigcup_\alpha^n G_\alpha$ is a finite subcover for $\mathbb{\hat{C}}$. (Convince yourself that if there were points left out, this would lead to a contradiction).   
