I need help with my proof in particular. I am aware that there is a similar question elsewhere. Can someone please verify my proof or offer suggestions for improvement?
Prove that every subset of $\mathbb{R}$ is compact in the finite complement topology.
Let $Y \subseteq \mathbb{R}$. If $Y$ is finite, the result is trivial. Suppose $Y$ contains infinitely many points. Let $\{A_\alpha\}_{\alpha \in J}$ be a collection of open sets of $\mathbb{R}$ which cover $Y$. If possible, pick an index $\beta \in J$ such that there exists a $y \in Y$ such that $y \notin A_\beta$. If such an index doesn't exist, any $\{A_\alpha\}$ would form a finite subcover of $Y$. If, on the other hand, such an index does exist, there are only finitely many elements $y_1, y_2, \ldots, y_n$ such that $y_i \notin A_\beta$ for each $1 \leq i \leq n$. Now, for each $1 \leq i \leq n$, we can pick an index $\gamma_i$ such that $y_i \in A_{\gamma_i}$. But then, $\{A_\beta, A_{\gamma_1}, \ldots, A_{\gamma_n}\}$ forms a finite subcover of $Y$.