Proving that $S^1$ is closed in $\mathbb{R}^2$ I want to prove that $S^1 = \{ (x,y): x^2+y^2=1\}$ is a closed subset in $\mathbb{R}^2$ in that following manner: I want to show that $(S^1)^c = R^2\setminus S^1$ is open. For this let $a=(a_1,a_2)\in (S^1)^c$ so $a_1^2+a_2^2>1$ or $a_1^2+a_2^2<1$. Let $a_1^2+a_2^2>1$. Let $r = d(a,S^1)$ and let's consider the open ball $B(a,r)$ and $x=(x_1,x_2)\in B(a,r)$. I'm stuck here to show that $x_1^2+x_2^2>1$. 
Any help? (Obs: $d(x,S) = \inf\{d(x,y): y\in S\}$)
 A: It would be easier if you note $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined by $(x, y) \mapsto x^2+y^2$ is continuous and $S^1 = f^{-1}( \{1\})$.
A: I will denote by $d$ the usual distance in $\mathbb{R}^2$. Let $A=(a_1,a_2)$, let $D=\frac A{\sqrt{{a_1}^2+{a_2}^2}}$, let $r=d(A,D)$ (it is equal to your $r$), let $B=(0,0)$ and let $C$ be an arbitrary point of $B(A,r)$. I will prove that $d(C,B)>1$. In order to that, note that $B$, $D$, and $A$ are colinear, with $D$ between $A$ and $B$; therefore, $d(B,A)=d(B,D)+d(D,A)=1+d(D,A)$. On the other hand, by the triangle inequality,$$d(B,A)\leqslant d(B,C)+d(C,A),$$and this is equivalent to$$1+d(D,A)\leqslant d(B,C)+d(C,A).$$Therefore\begin{align}d(C,B)&=d(B,C)\\&\geqslant1+d(D,A)-d(C,A)\\&=1+r-d(C,A)\text{ (because $C\in B(A,r)$)}\\&>1.\end{align}
A: An alternative way. Consider the map from the plane to the line given by $x^2+y^2$, and show that it's continuous. The circle is just the inverse image of $1$ under this continuous map and hence is closed. 
A: Here are two random arguments for fun, I think that they work.
Identify $\mathbb C$ with $\mathbb R^2$. Note that $[0,2 \pi]$ is closed in $\mathbb R^1$. Endow it with the subspace topology.
Consider  $f:[0,2 \pi]  \to \mathbb C$ given by $f(x)=e^{ix}$. Notice that its image is $S^1 \subset \mathbb C$.


*

*this is a continuous map from a compact space to a Hausdorff space, and hence closed, so $S^1$ is closed.

*Since $f(x)=f(y) \iff x=y$ or $x,y=0,1$. In other words, this map identifies the points $0,2 \pi \in [0,2 \pi]$. But then this induces a map $\tilde{f}:[0,2 \pi]/{\sim} \to S^1 \subset \mathbb C$, where $x \sim y \iff x,y=0,1$. Indeed, $\tilde{f}$ is a homeomorphism, and so $S^1$ is closed in $\mathbb C$.
A: Use the triangle inequality :$ \lvert x+y\rvert \le \lvert x\rvert +\lvert y\rvert $, in the form $\lvert x-y \rvert\ge \lvert x \rvert -\lvert  y \rvert $.  Use it on the triangle formed by $x ,a $ and $(0,0) $...
Note:  $ \lvert x- a\rvert   \lt r $.  And $\lvert a\rvert = 1+r $.  Therefore,  $\lvert x \rvert =\lvert a-(a-x)\rvert \ge \lvert a\rvert -\lvert a-x\rvert  \gt 1+r-r=1$...
A: To prove that $(S^1)^c$ is open it suffices that for any $a\in (S^1)^c$ there exists some $z>0$ such that $B(a,z)\cap S^1$ is empty. We do not need to take $z=d(a,S^1).$ 
Let $a=(a_1,a_2)$ with $a_1^2+a_2^2=1+s\ne 1$.  
If $z>0$ and  $z$ is small enough that $\max (|a_1z|,|a_2z|, |z|^2)<|s|/6$ then for any $(a_1+d_1,a_2+d_2)\in B(a,z)$ we have $|d_1|<z$ and $|d_2|<z,$ $$\text { so }\quad |2a_1d_1+2a_2d_2+d_1^2+d_2^2|<|s|$$  $$\text {so } \quad  (a_1+d_1)^2+(a_2+d_2)^2=(1+s)+(2a_1d_1+2a_2d_2+a_1^2+d_2^2)\ne 1.$$
