Prove: The set $G = \{ X \in\mathbb R^2 : d(X,0) \le 1\}$ is not open in $\mathbb R^2$. (Here $X=(x,y), 0=(0,0)$)
We will attempt to prove G is not open in $R^2$ by contradiction.
Let $x=1, y=0$ then by the restrictions imposed on $G$:
$d(X, 0) = \sqrt{(1 - 0)^2 + (0 - 0)^2} = \sqrt{1} = 1$
Thus $(1, 0) \in G$. Let $a = (1, 0)$.
Let us assume $G$ is open, by definition, there exists an $\epsilon > 0$ such that $B_\epsilon(a) \in G$.
Let $\epsilon > 0, r > 0$ and $r < \epsilon$ then $(1 + r,0) \in B_\epsilon(a)$. Let $b = (1 + r, 0)$.
Again by the restrictions imposed on G:
$d(b, 0) = \sqrt{((1 + r) - 0)^2 + (0 - 0)^2}$ = $\sqrt{(1 + r)^2}$ = $1 + r$.
Since $r > 0$ then $d(b, 0) > 1$ by the restrictions of $G$, $b \notin G, B_\epsilon(a) ∉ G$.
This is a contradiction meaning $G$ is not open in $\mathbb R^2$.
Here is an easier approach:
The map $f: \mathbb{R}^2 \to \mathbb{R}: X \mapsto d(X,0)$ is continuous. Consequently, the set $G= f^{-1}(]-\infty,1])$ is a non-empty, proper closed subset of $\Bbb{R}^2$. Since $\Bbb{R}^2$ is connected, $G$ can't be open.