Let $\tau:=\{A\subseteq \mathbb R: \mathbb R\setminus A \operatorname{finite}\}$
Show that $[0,1]$ is compact.
Let $(A_{n})_{n}$ be an open cover of $[0,1]$
I had two ideas:
$1.$ Since $[0,1]\subseteq\bigcup_{n \in \mathbb N}A_{n}$ there exists an $x \in [0,1]$ and an $\alpha \in \mathbb N$ so that $x \in A_{\alpha}$
I want to claim that since $A_{\alpha}\in\tau\Rightarrow\mathbb R\setminus A_{\alpha}$ is finite and thus $[0,1]\subseteq A_{\alpha}$ but I have not been able to show this directly.
Then I came to second idea:
$2.$ By contradicton, assume that the $A_{\alpha}$ chosen in $1.$ does not contain all of $[0,1]$ and since $\mathbb R \setminus A_{\alpha}$ is finite, $[0,1]$ is at most not covered by $A_{\alpha}$ by finitely many points, and thus we can find a finite subcover.
Any ideas/suggestions/corrections, are well-received.
Let $(A_\alpha)$ be an open cover of $[0,1]$. In particular there exists $A_\beta$ such that $0\in A_\beta$. Since $A_\beta$ is open, $A_\beta$ covers all but finitely many points ${x_1, \dotsc, x_n}$ of $[0,1]$. The remaining finite number of points not covered can be covered by choosing $A_{\beta_{i}}$ such that $x_i\in A_{\beta_{i}}$.