prove that the unit circle $x^2+y^2=1$ is a closed set I need to prove that the unit circle $x^2+y^2=1$ is a closed set in $\mathbb{R}^2$ is closed using convergent sequence method.
 A: Consider the set $S=\{(x,y)\in\mathbb{R}^2\;|\; x^2+y^2=1\}$. Consider a sequence $(x_n,y_n)$ in S, s.t. it converges in $\mathbb{R}^2$.  We have $(x_n,y_n)\rightarrow (x,y)\in\mathbb{R}^2$ iff $x_n\rightarrow x\in\mathbb{R}$ and $y_n\rightarrow y\in\mathbb{R}$. This means that $x_n+y_n\rightarrow x+y$. Since $(x_n,y_n)\in S$ for all $n\in\mathbb{N}$ we have $x_n^2+y_n^2=1$ for all $n\in\mathbb{N}$. Therefore $x_n^2+y_n^2$ is a constant sequence (given equality to 1). Since $x_n^2+y_n^2\rightarrow x^2+y^2$ we get $x^2+y^2=1$. Therefore $(x,y)\in S$. This completes our proof. 
A: HINT: Work in polar coordinates: then a sequence in $S^1$ is of the form $\big\langle\langle 1,\theta_n\rangle:n\in\Bbb N\big\rangle$, and you can think about it in terms of the sequence $\langle\theta_n:n\in\Bbb N\rangle$, which you can take to be in $[0,2\pi]$. (This isn’t the only way, but it makes life pretty easy.)
Added: Note that if the sequence converges to the point whose rectangular coordinates are $\langle 1,0\rangle$, this won’t quite work, since you may have infinitely many angles a little larger than $0$ and infinitely many a little less than $2\pi$. In that case use angles in the range $[-\pi,\pi]$ instead. 
A: The euclidean norm $$(x,y)\mapsto ||(x,y)||$$
is continuous then if $(x_n,y_n)\in S^1\to(x,y)$ we have $$||(x_n,y_n)||=1\to ||(x,y)||=1$$
so $$(x,y)\in S^1\quad \text{hence} \quad S^1\quad \text{is closed subset}$$
A: Another way is to show that the complement is open. The complement of the your set is
$$S = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 \neq 1\}$$
Consider $(x,y) \in S$. For this $(x,y)$, the $\epsilon$-ball centered at $(x,y)$, where $\epsilon = \left\vert \sqrt{x^2+y^2} - 1 \right\vert$, lies completely in $S$. Hence, $S$ is open.

EDIT:
Here is a pictorial description of finding the open sets. The blue colored unit circle if your set $x^2+y^2 = 1$. The remaining white space is the complement of the unit circle. You want to show that this complement is open.
Now pick a point, say $(0.25,0.8)$. This is a point "inside" the blue unit circle circle. For this point, you can find a open ball of radius $1-\sqrt{0.25^2 + 0.8^2}$, which lies completely "inside" the blue unit circle and therefore lies completely in the complement of the unit circle.
Similarly, pick a point, say $(1.125,0.6)$. This is a point "outside" the blue unit circle circle. For this point, you can find a open ball of radius $\sqrt{1.125^2 + 0.6^2}-1$, which lies completely "outside" the blue unit circle and therefore lies completely in the complement of the unit circle.
In general, you can do this for any point not lying on the unit circle. This shows that the complement of the unit circle is open.  

