In a normed vector space $E$ the only open and closed set is $E$ and the empty set as a corollary to connected properties. Also an closed ball cannot be open, I would like to prove this result with only the fact that :
A set $O$ of a normed vector space $(E,\Vert\Vert)$ is open if and only if $$\forall x\in O\quad \exists\varepsilon>0\quad B(x,\varepsilon) \subset O.$$
So if the closed ball $B(a,r]=O$ is open whenever I choose an element of $O$ there exist $\varepsilon>0$ such that $B(x,\varepsilon)\subset O$. With a picture is cleary false by choosing an element $x$ such that $\Vert x-a \Vert=r$. I have to choose $x=a+r\nu$ where $\nu$ is an unitary vector. Then I have to prove that the open ball centered at $a+r\nu$ with a strictly positive radius $\varepsilon$ cannot be included in $O$.
Am I right?