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 $(\frac{\pi}{2},\frac{3\pi}{2})$.

Yes, that is right. And $U=\{x\in\mathbb{R}^n\colon\|x\|>0\}\cap X$.

A more fancy approach: it will suffice to say that the singleton $\{0\}$ is a clopen (=closed and open) set. It is closed, because one point sets are always closed. It is open, because it is $X \cap \{ x \in \mathbb{R}^n \ : \ || x || < 1 \}$

Alternative: the norm $\|\cdot \|$ is 1-Lipschitz by the triangular inequality, whence continuous. The image of your set under this continuous map is $\{0\}\cup\{\pi\}$, which is disconnected. So your set can't be connected, since the continuous image of a connected space is connected.

Any open set O in $\mathbb R^n$ intersected with $X$ yields an open set $O\cap X$ in X.

Proof: by definition, a set $S\subseteq X$ is open in $X$, iff for each $x \in S$ an $\epsilon>0$ exists such that the ball $B(x,X,\epsilon)= \{ y \in X : ||x-y||< \epsilon\} \subseteq S$. Now if $S=O\cap X$ for an open set $O \subseteq \mathbb R^n$, then for each $x \in S$, there exists an $\epsilon$ such that $B(x,\mathbb R^n,\epsilon) \subseteq O$. This implies that $B(x,X,\epsilon)=B(x,\mathbb R^n,\epsilon)\cap X \subseteq O\cap X=S$.

So $V$ is open. Similarly, $U= \{x\in \mathbb R^n: 3<||x||<4\}\cap X$ is also open in $X$. As $X$ is the disjoint union of two open sets, it is not connected.