Prove subset of $\mathbb R^2$ is open Problem. Let $G = \{(x,y): x \ne y\}$. Prove $G$ is an open subset of $\mathbb R^2$.

What I am thinking: If I could show that $\mathbb R^2 \setminus G = \{(x,y): x = y\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?
Let the function $f\colon \mathbb R^2\rightarrow \mathbb R,$ $(x,y)\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have 
$$G=f^{-1}(\mathbb R^*)$$
Do you know what theorem we should use to conclude?
HINT: $\Bbb{R}^2 - G$ results in a line. Can this complete $G$?