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.
Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.
Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.
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.