prove that $s=\left \{ x:\left | x \right |\geqslant 1\left. \right \} \right.\subseteq \mathbb{R}$ not open

Question

prove that $s=\left \{ x:\left | x \right |\geqslant 1\left. \right \} \right.\subseteq \mathbb{R}$

I thought that we take a point at the one of the boundray lines $x=1$ or $x=2$ and show that this point is not interior.

i don't know how to do so

I'm a high school student so please don't mock me

Answer 1

If you take $x=1$ this point is in $s$. A ball centered in $1$ with radius $r$ have the form: $$B(1,r)=(1-r,1+r)$$ For any radius $r$ the points in the interval with $x<1$ are not in $s$ then $x=1$ is not an interior point of $s$ and then $s$ is not open. (Or for any $r$ the points in $(1-r,1) \subset (1-r,1+r)$ are not in $s$)