Properties of this topology on $\Bbb X$.

For given the usual topology $$\tau$$ on $$\Bbb{R}$$, define the compact complement topology on $$\mathbb{R}$$ to be $$\tau'=\{A\subseteq \Bbb{R}:A^C\text{ is compact in }(\Bbb{R},\tau)\} \cup \{\emptyset \}.$$

Hausdorff : let $$G_1$$ and $$G_2$$ be disjoint open sets containing $$x$$ and $$y$$ for $$x \ne y$$ then $$G_1 \cap G_2 = \varnothing$$ $$G_1^C \cup G_2^C = \Bbb R$$ but $$\Bbb R$$ is not compact. So space is not Hausdorff.

Connectedness: Will the above also work for connectedness? Showing it's connected?

And are there any easy proof on compactness? I already saw this link but can't understand.

Compactness is not too hard: suppose $$\{U_i: i \in I\}$$ is an open cover of $$\mathbb{R}$$ by non-empty open sets. Take any $$U_{i_0}$$ in this cover. Then $$U_{i_0}^c$$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $$F \subseteq I\setminus \{i_0\}$$ that covers $$U_{i_0}^c$$ and then $$\{U_i: i \in F \cup \{i_0\}\}$$ is a finite subcover of our original cover. Hence the space is compact.