I have the following problem: The set $\mathbb{R}^2$ has a topology T for which the closed sets are the empty set and the finite unions of vector subspaces. Let X denote the set $\mathbb{R}^2 - \{0\}$ with the subspace topology induced by T . By considering the subspace topology on $S_1$ ⊂ $\mathbb{R}^2$, or otherwise, show that X is compact.
Is it valid to say: The open sets are just $\mathbb{R}^2$- {finite unions of vector spaces}. So take an open cover, pick an arbitrary element in the cover. Then this covers all but a finite union of vector subspaces, which can then be covered by a finite number of open sets in the cover.
I'm doubtful, since the answer seems too short for the context the question is in, and doesn't use the hint given.
Observation 1: $S^1$ in the usual topology is compact. So it is also compact in this weaker topology (vector subspaces are already usual-closed, so open sets in this topology are usual-open sets). Any open cover in the weaker topology is one in the usual topology etc.
Observation 2: if two vector subspaces intersect in $X$, they also intersect in $S^1$: a have a non-zero (because $0 \notin X$) vector $v \in V_1 \cap V_2$, then $\frac{1}{\|v\|}\cdot v \in V_1 \cap V_2 \cap S^1$. This in fact holds for any intersection of vector subspaces in $X$.
From 2 follows that a family of closed sets in $X$ has the FIP iff its intersections with $S^1$ have the FIP there.
Now apply the closed sets (or FIP family) formulation of compactness to see $X$ is compact.