Showing that $\{x\in\mathbb R^n: \|x\|=\pi\}\cup\{0\}$ is not connected I do have problems with connected sets so I got the following exercise:

$X:=\{x\in\mathbb{R}^n: \|x\|=\pi\}\cup\{0\}\subset\mathbb{R}^n$. Why is $X$ not connected?

My attempt: I have to find disjoint open sets $U,V\ne\emptyset$ such that $U\cup V=X$ .
Let $U=\{x\in\mathbb{R}^n: \|x\|=\pi\}$ and $V=\{0\}$. Then $V$ is relative open since $$V=\{x\in\mathbb{R}^n:\|x\|<1\}\cap X$$ and $\{x\in\mathbb{R}^n:\|x\|<1\}$ is open in $\mathbb{R}^n$.
Is this right? and why is $U$ open?

Yup, that's right! And similarly, you can see that $U$ is open because $U=W\cap X$ where $$W=\{x\in\mathbb{R}^n:\tfrac{\pi}{2}<\|x\|<\tfrac{3\pi}{2}\}$$
is an open set of $\mathbb{R}^n$. One way of seeing that $W$ is open is noting that the function $\|\cdot\|:\mathbb{R}^n\to\mathbb{R}$ is continuous, and $W$ is the preimage of the open set $(\frac{\pi}{2},\frac{3\pi}{2})$.
