Show that in the finite-complement topology on $\mathbb{R}$, every subspace is compact.
We can generalize the problem
$(X,\mathcal{T}_{\text{cofinite}})$ and $Y\subseteq X$ $\Rightarrow$ $Y$ is compact.
Proof: $(X,\mathcal{T}_{\text{cofinite}})$. $Y\subseteq X$. Let $U=\{ U_\alpha \in \mathcal{T}_X |\alpha \in J$ and $U_\alpha \neq \emptyset \}$ be an open cover of $Y$ in $X$, i.e. $Y\subseteq \bigcup_{\alpha \in J}U_\alpha$. Since $U_\alpha \in \mathcal{T}_X$ $\forall \alpha \in J$, either $X-U_\alpha$ is finite or $U_\alpha =\emptyset$. $U_\alpha \neq \emptyset$, $\forall \in J$. So $X-U_\alpha$ is finite, $\forall \alpha \in J$. If $Y=\emptyset$, then $Y$ is compact trivially. Assume $Y\neq \emptyset$. Since $Y\neq \emptyset$, $\exists y\in Y$. $y\in U_{\alpha_0}$, for some $\alpha_0 \in J$. Let $Z=\{ z\in Y|z\notin U_{\alpha_0}\} \subseteq X-U_{\alpha_0}$. Since $X-U_{\alpha_0}$ is finite, $|Z|=n\in \Bbb{N}$. So $Z=\{z_1,..,z_n\}\subseteq X-U_{\alpha_0}$. $z_k \in U_{\alpha_k}$, $\forall 1\leq k\leq n$. Thus $\{ U_{\alpha_0}, U_{\alpha_1},…, U_{\alpha_n}\}$ is a finite subcover of $U$. Hence $Y$ is compact. Is this proof correct?
Potential approach(2) is to use theorem 26.9 definition of compactness.