showing that some topology of $\mathbb{R}^2$ is not Hausdorff Let $T$ be a topology on $\mathbb{R}^2$ generated by a basis consisting of sets that equal $\mathbb{R}^2$ subtracted with finitely many complete lines.
The question is that showing that $T$ is not Hausdorff.
In fact, I solve this problem using connectedness of $\mathbb{R}^2$. But I am curious whether other more ingenuine approaches to this problem exist.
 A: Suppose that $U$ and $V$ are disjoint basic open sets. Then there are finite sets $\mathscr{L}_U$ and $\mathscr{L}_V$ of lines such that $U=\Bbb R^2\setminus\bigcup\mathscr{L}_U$ and $V=\Bbb R^2\setminus\bigcup\mathscr{L}_V$, and
$$\begin{align*}
\varnothing&=U\cap V\\
&=\left(\Bbb R^2\setminus\bigcup\mathscr{L}_U\right)\cap\left(\Bbb R^2\setminus\bigcup\mathscr{L}_V\right)\\
&=\Bbb R^2\setminus\left(\bigcup\mathscr{L}_U\cup\bigcup\mathscr{L}_V\right)\\
&=\Bbb R^2\setminus\bigcup(\mathscr{L}_U\cup\mathscr{L}_V)\;,
\end{align*}$$
and therefore $\bigcup(\mathscr{L}_U\cup\mathscr{L}_V)=\Bbb R^2$, i.e., $\Bbb R^2$ is the union of finitely many lines. But this is impossible: the complement of each line is a dense open set in $\Bbb R^2$ in the usual topology, and it’s an easy exercise to show that the intersection of finitely many dense open sets is dense and therefore non-empty. (In fact $\Bbb R^2$ isn’t even the union of countably many lines. Each line is a closed, nowhere dense subset of $\Bbb R^2$ in its usual topology, and by the Baire categor theorem $\Bbb R^2$ with its usual topology is not the union of countably many closed, nowhere dense sets.)
Thus, $\Bbb R^2$ in this new topology does not contain any disjoint non-empty open sets and therefore cannot be Hausdorff.
Added: Let $\tau$ be this new topology, and suppose that $\langle\Bbb R^2,\tau\rangle$ were Hausdorff; then every subspace of $\langle\Bbb R^2,\tau\rangle$ would also be Hausdorff. Let $Y$ be the $x$-axis, and let $\tau_Y$ be the subspace topology on $Y$; then it’s clear that $\tau_Y$ is the cofinite topology on $Y$, which is $T_1$ but not Hausdorff. Thus, $\langle\Bbb R^2,\tau\rangle$ is not Hausdorff.
