The set $H=\{(x,y)\in \Bbb{R^2}:\;3x+2y=5 \},$ is a closed subset of $\Bbb{R^2}$

Looks Good! Alternatively, take $\alpha=(x,y) \in \Bbb R^2 \setminus H$. Take $r=d(\alpha,H)>0$, the distance from the point $\alpha$ to the line $H$. Then $$B(\alpha,r) \subset \Bbb R^2 \setminus H$$ showing $\Bbb R^2 \setminus H$ is open.