Is the set $\{(x,y)\in \mathbb{R}^2 \mid x^2+y^2 \neq 1\}$ open in $\mathbb{R}^2$? Consider $S :=\{(x,y)\in \mathbb{R}^2 \mid x^2+y^2 \neq 1\}$. Show that $S$ is open in $\mathbb{R}^2$.
My attempt: We see that $S$ can be  written as union of two sets $U:= \{(x,y) \in \mathbb{R}^2 \mid x^2+y^2 <1\}$ and $V:=\{(x,y)\in \mathbb{R}^2\mid x^2+y^2>1\}$. We will show that $U$ and $V$ are open in $\mathbb{R}^2$.
$\fbox{$U$ is open set:}$ Let $(x,y)\in U$. Hence $d((x,y), (0,0)) <1$. Define $r:=\frac{1}{2}[1-d((x,y),(0,0))]$. Since  $d((x,y),(0,0))<1$, we get $r>0$. Let $d$ be the standard Euclidean metric. Thus, for  $(x,y)\in U$ there exists $r>0$ such that $B_d((x,y),r) \subset U$. So, $U$ is open in $\mathbb{R}^2$.
$\fbox{$V$ is open set:}$ Let $(x,y)\in V$. Hence $d((x,y), (0,0)) >1$. Define $s:=\frac{1}{2}[d((x,y),(0,0))-1]$. Since  $d((x,y),(0,0))>1$, we get $s>0$. Let $d$ be the standard Euclidean metric. Thus, for  $(x,y)\in V$ there exists $s>0$ such that $B_d((x,y),s) \subset V$. So, $V$ is open in $\mathbb{R}^2$.
Since $S = U\cup V$, we get $S$ is open in $\mathbb{R}^2$.
Is my answer correct?
 A: The logic of the proof is correct, but it reads like word salad. For example,

Let $d$ be the standard Euclidean metric.

You've put that in twice, and after you've already used $d$. You're considering a metric space so one assumes that you already have a metric on $\mathbb{R}^2$. You even use the properties of $d$ before the above.

Thus, for  $(x,y)\in U$ there exists $r>0$ such that $B_d((x,y),r) \subset U$. So, $U$ is open in $\mathbb{R}^2$.

Why "thus"? Why do you repeat "for $(x,y)\in U$" like that? You've already said that $(x,y)$ is in $U$ at the beginning. Anyway here is my suggested rewrite of part of the proof:

Define $C=\left\{(x,y)|x^2+y^2=1\right\}$, $O=(0,0)$.


$V$ is open: Let $X\equiv(x,y)\in V$, then $d(X, O) >1$ by definition. Let $s=\frac{1}{2}(d(X,O)-1)$. Since $d(X,O)>1$, $s>0$. Let $B\equiv  B_d(X,s)$. If $q\in B$ then $d(q,O)+s>d(q,O)+d(q,X)\geq d(X,O)$ and so $$d(q,O)\geq d(X,O)-s=\frac12(d(X,O)+1)>1,$$ hence $q\not\in C$, thus $B \cap C=\emptyset$, hence $B \subset V$. So, $V$ is open in $\mathbb{R}^2$.

A: Yes, that's perfect. Another approach would be using that the preimage of a regular value through a continuous function is closed. Just take $f(x,y)=x^2+y^2$, since $\mathbb{S}^1=f^{-1}(\{1\})$, the complementary( your set $S$) is open
