Show that in the finite complement topology on $\mathbb{R}$, every subspace is compact I need to show two things:
1) Show that in the finite complement topology on$\mathbb{R}$, every subspace is compact
2) If $\mathbb{R}$ has the topology consisting of sets $A$ such that $\mathbb{R}-A$ is either countable or all of $\mathbb{R}$, is $[0,1]$ a compact subspace?
For 1) I tried to think like this: suppose $Y$ a subspace of $\mathbb{R}$ with the finite complement topology. Suppose then, an open cover for $Y$. Since it's open, the complement of $Y$ is finite. I thought I'd come up with something here but I don't think it's correct
For 2) I've found the example $A_n = [0,1]-\{\frac{1}{n},\frac{1}{n+1},\cdots\}$ but I don't know how it answers the question. Why this set cannot be covered by finitely many open subspaces?
 A: 
Take any open cover $U$ of a co-finite topology on a set $X$ .  Then for any $u\in U,$ we have $X\setminus u$ is finite. So, let $X\setminus u=\{x_1, x_2, \cdots, x_n\}$.
  Choose $u_1, u_2, \cdots, u_n\in U$ be such that $x_k\in u_k$ for each $k=1, 2, \cdots ,n.$
  Now $\{u,u_1, u_2, \cdots, u_n\}$ is a finite sub-cover of $X$ extract from an arbitrary open cover $U.$
  Hence $X$ is compact.

In your first question choose $X$ as a subspace of $\Bbb{R}$ with the co-finite topology.
For the second question note that rationals in $[0,1]$ are countable but not finite.
Let $\{q_1, q_2, q_3,  \cdots \}$ be enumeration of rational in $[0,1].$
Choose $A_n=\Bbb{R}\setminus(\Bbb{Q}\cap[0, 1])\cup\{q_n\}.$ Then the collection $(A_n)_{n\in\Bbb{N}}$ is a countable open cover of $[0,1]$  in the co-countable topology of $\Bbb{R}$ which has no finite sub-cover. 
A: Let X be a subspace of R in the finite complement topology, and let {Uα} be an open
cover for X. Pick a particular Uα0 = R− Aα0
, where Aα0
is finite. Then X − Uα0 ⊂ Aα0
is finite, say X − Uα0 = {a1, a2, . . . , an}. For each ai
, we can find an Uαi
that contains
it, so {Uα0
, Uαi
, . . . , Uαn
} is a finite subcover.
