Show that any line in $\Bbb R^2$ is closed Show that any line in $\Bbb R^2$ is closed under the usual topology
I think showing the complement is open is best way. 
Let $ax+by+c=0$ be a line in $\Bbb R^2$ as a line it is the set of $(x,y) \in \Bbb R^2 $ such that it satisfies the above equation for constants $a,b,c$ for simplicity call the set $L$ 
Take $(x,y) \in L^c$, the distance from $(x,y)$ to $L$ is $\displaystyle \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$
Let $\epsilon = \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$ then take as $U$ the ball $((x,y), \epsilon)$ then $U \cap L = \emptyset$ so $L^c$ is open.

The line defined by the equation $ax+by+c=0$  is the inverse image of the closed subset $0$ in $\mathbf R$, by the (continuous) map  $\begin{aligned}[t]\mathbf R^2&\longrightarrow \mathbf R,\\ (x,y)&\longmapsto ax+by+c.\end{aligned}$
