Hawaiian gradient Let $X$ be the subespace of $(R^2,τ_u)$ defined by
$X$=$\cup_{n=1}^∞\left\{{(x,y):(x-\frac{1}{n})^2+y^2= (\frac{1}{n})^2}\right\}$
Prove that $X$ is compact and connected

To see that it is compact, we have to check that it is closed and bounded.
For closed we have to prove that $\mathbb{R}-X$ is open, i think it is trivial because all element of $X$ are circles and we can take any element of $\mathbb{R}-X$ and get an open ball centered on that element and contained in $\mathbb{R}-X$ but I don´t know how to formalize it.
For bounded we can contain it in the square $[0,2]$ x $[-1,1]$. how can I write it formal? It´s enough to say that $(\frac{1}{n})^2$ < 1 and the center is $(\frac{1}{n},0)$? how can I write it better?
To see that it is connected, I have no idea.
 A: I assume that $(R^2,τ_u)$ is the space $\Bbb R^2$ endowed with the usual topology.
The set $X$ is a union of circles $X_n=\{(x,y)\in\Bbb R^2:(x-\frac{1}{n})^2+y^2= (\frac{1}{n})^2\}$ with the maximal radius $1$ passing through a point $0=(0,0)$. Therefore a distance from any point of $x$ to $0$ is a most $1$ and so the set $X$ is contained in a disk of radius $1$ centered at $0$, so the set $X$ is bounded.
Let $x\in\Bbb R^2\setminus X$ be any point. Since $0\in X$, $x\ne 0$, and so the distance $d$ from $x$ to $0$ is positive. Let $O_x$ be an open ball of radius $d/2$ centered at $x$. It easy to check that $O_x\cap X_n=\varnothing$  for each $n>2/d$. Since each $X_n$ is compact, it is closed in $\Bbb R^2$,  so $O_x\setminus\bigcup_{n\le  2/dn} X_n$ is an open neighborhood of $x$ disjoint with $X$. So the set $X$ is closed in $\Bbb R^2$.
The set $X$ is connected being a union of a family of connected sets $X_n$ with a common point $z$, because for each partition of $X$ into two closed and open subsets, the partition set containing $z$, contains all $X_n$.
