In a metric space,the union of two open sets is open,i.e, $U=(4,6)\cup (9,14)$.
1)Is this true because all the points that belong to $U$ are interior points?
But $U$ is not an open ball.
2)Is it because there is not open ball with center in $U$ that contains only interior points? 3)Could you consider the balls with center outside U to prove that not all points are interior?
To answer your questions in order:
1) Yes. You may pick any $u\in U$, and note that you can construct an $\epsilon=(1/2)\min\{|6-u|,|4-u|,|9-u|,|14-u|\}$, and then the open ball $B(u,\epsilon)$ is contained in $U$.
2) Yes. If $U=B(x,r)$ for some $x\in U$ and $r>0$, then by considering cases where $x\in (4,6)$ or $x\in (9,14)$, and using the definition of an open ball $B(x,r)=\{z\in\mathbb{R}:d(z,x)<r\}$, one can show that the points in $(4,14)\setminus(4,6)\cup (9,14)$ are in $B(x,r)$.
3) If there is a ball with centre outside of $U$, say $B(y,r')$ for some $y\in\mathbb{R}\setminus U$, then by the definition of an open ball, you can show that $B(y,r')\neq U$.
