Show that interval $(a, b)$ is not open in $\mathbb{R}^2$ I know that interval $(a, b)$ is open in $\mathbb{R}$. To show that interval $(a,b)$ is open in $\mathbb{R}$, I have done so: Let it be $x\in (a,b)$. Enough to find an open ball containing the point $x$, and that is included in the interval $(a,b)$. Suffice to get $0<\epsilon\leq \min \{\vert b-x\vert, \vert a-x \vert\}$. In this case $D(x,\epsilon)$ containing the point $x$ and $D(x,\epsilon)\subseteq (a,b)$. I do not know a good act. If it is that I did very well then correct, but I do not know how to prove this fact: To show that interval $(a,b)$ is not open in $\mathbb{R^2}.$
Please if anyone has the opportunity to help, to make verification of keti example, thank you preliminarily

Hint $x + \frac{\epsilon}{2}(0,1) \in D_\epsilon (x)$...
As $x + \frac{\epsilon}{2}(0,1)$ has a non-zero $y$-coordinate, it is not in $(a,b)$.
P.S. If $a,b$ are not real numbers, but points in $\mathbb R^2$, you can do the same by replacing $(0,1)$ by an unit vector $u$ such that $u \perp {ab}$.
