The union of circles with center $(1/n,0)$ and radius $1/n$ is closed in $\mathbb{R}^2$ 
Let $C_n$ be a circle of radius $\frac{1}{n}$ and centered at $\left(\frac{1}{n},0\right)$ for all $n \geq 1$. Then the set $$ X = \bigcup\limits_{n=1}^{\infty}C_n$$ is closed as a subset of $\mathbb{R}^2$.

I tried to prove this using the usual tactic.
$X$ is closed if and only if its complement is open. The complement of $X$ is the set $$X^c = \bigcap\limits_{n=1}^{\infty}\mathbb{R}^2 \setminus C_n .$$ Now $$\mathbb{R}^2\setminus C_n = \left\{(x,y) \in \mathbb{R}^2 \left\vert \left(x - \frac{1}{n}\right)^2 + y^2 \neq \frac{1}{n^2}\right.\right\} .$$ Now, the set $$\left\{(x,y) \in \mathbb{R}^2 \left\vert \left(x - \frac{1}{n}\right)^2 + y^2 < \frac{1}{n^2}\right.\right\}$$ is open as for any $(x,y)$ in this set, we can find a open disc of radius $\varepsilon > 0$ which is contained entirely in this set. Similarly, we can show that the set $$\left\{(x,y) \in \mathbb{R}^2 \left\vert \left(x - \frac{1}{n}\right)^2 + y^2 > \frac{1}{n^2}\right.\right\}$$ is open. 
But arbritrary intersections of open sets is not open. I am not sure how I can conclude that the set is closed ? 
The $X$ looks like a union of boundary sets which I know are closed. But then again, arbritrary union of closed sets are not closed. I am not sure how to proceed. Thanks in advance for your help. 
 A: Denote by ${\rm int}(C_n)$, resp., ${\rm ext}(C_n)$ the open interior, resp. open exterior of the disk bounded by $C_n$. Then the complement of your set $X$ can be written as
$$X^c={\rm ext}(C_1)\cup\bigcup_{n=1}^\infty\bigl({\rm int}(C_n)\cap{\rm ext}(C_{n+1})\bigr)\ .$$
As ${\rm int}(C_n)\cap{\rm ext}(C_{n+1})$ is open for all $n$ the set $X^c$ is open.
A: Take X(n) as the union of the circles from 1 to n, and Y(n) as the union of the rest - than X is the union of the closed set X(n) and the set Y(n);
take a point (x,y) outside of X -> either /x/ or /y/ (or both) is larger than 0;
take N such that 1/N is smaller than 0.25/x/ or 0.25/y/;
show that the distance to each circle of Y(N) is larger than 1/N;
show that the distance to X(N) is positive (X(N) is closed);
-> the distance to X is positive
-> X is closed
A: Let $B(r)$ be the open disk of radius $r$. Then notice that for each $n$
$$
X\setminus B\left(\tfrac1n\right)=\bigcup_{k=1}^{2n}\left[C_k\setminus B\left(\tfrac1n\right)\right]
$$
which is closed since it is the finite union of closed sets.
Suppose that $X$ is not closed, then there is some $x\not\in X$ that is a limit point of $X$. Since $X\setminus B\left(\frac1n\right)$ is closed, we must have $x\in B\left(\frac1n\right)$ for each $n$. That is,
$$
x\in\bigcap_{n=1}^\infty B\left(\tfrac1n\right)=\{(0,0)\}
$$
However, $x=(0,0)\in C_1\subset X$; contradiction. Therefore, $X$ is closed.

Another way to look at this, is that $X^C$ is the union of open sets:
$$
X^C=\bigcup_{n=1}^\infty{\underbrace{\left[\,\overline{B\left(\tfrac1n\right)}\,\cup\,\bigcup_{k=1}^nC_k\right]}_{\substack{\text{union of finitely many closed}\\\text{sets; therefore, closed}}}}^C
$$
Therefore, $X$ is closed.
A: If you are not limited to use certain things: 
Theorem. Let $C_{k}$ be a set closed in $\mathbb{R}^{n}$ for all $k \geq 1$; let $C := \bigcup_{k\geq 1}C_{k}$ be locally finite (i.e. for every $x \in C$ there is some $r > 0$ such that for the open ball $V^{x}(r)$ of center $x$ and radius $r$ we have $V^{x}(r) \cap C_{k} \neq \varnothing$ for only finitely many $k$); then $C$ is closed in $\mathbb{R}^{n}$.
Things get easy when applying the theorem: if
$C_{k} := \Big\{ (x,y) \in \mathbb{R}^{2} \mid \Big( x - \frac{1}{k} \Big)^{2} + y^{2} = \frac{1}{k} \Big\}$ for all $k \geq 1$, then $C_{k}$ is the preimage of the singleton $\{ \frac{1}{k} \}$ under the map $(x,y) \mapsto (x-\frac{1}{k})^{2} + y^{2}$ and hence is closed in $\mathbb{R}^{2}$ for all $k \geq 1$; moreover, it is easy to see that $C$ is locally finite; so by the theorem we conclude that $C$ is also closed in $\mathbb{R}^{2}$.
