In $\mathbb {R}^{2}$ show that $\big\lbrace\left(x,\,0\right)\mid\;-1I can show that it is not closed be cause the complement $(-\infty ,1]\cup [1,\infty )$ is not open, but I can't prove that it is not open. 
Anyone have any hint ?
 A: Consider the set:
$$S_e=\{(x,y)|-1\le x \lt 1,-e\le y\le e\}$$
for some $e\ge 0$.
For any $e\gt0$ the set is open, because we can find an open 
disc neighbourhood for every point in it.
If $e=0$ then there is no open disc neighbourhood in $\mathbb R^2$.
A: Geometrically we can see like this , observe that there exists a neighborhood such that it contains points other than the set you gave. 
Can you figure it out now?

It may help!
A: I think you need to recall the following important definitions.
Definition 1. For $(x,y)\in\Bbb R^2$ and $\epsilon >0$, we write
$$B_\epsilon(x,y)=\bigg\{(a,b)\in\Bbb R^2:\sqrt{(x-a)^2+(y-b)^2}<\epsilon\bigg\}.$$ The set $B_\epsilon(x,y)$ is called open disc with center $(x,y)$ and radius $\epsilon$.
Definition 2. Let $A\subset\Bbb R^2$. Then $A$ is open if for each $(x,y)\in A$ there exists $\epsilon>0$ such that $B_\epsilon(x,y)\subset A$.
Equivalently, we get
Definition 3. Let $A\subset\Bbb R^2$. Then $A$ is not open if there exists $(x,y)\in A$ such that for all $\epsilon>0$, we have $B_\epsilon(x,y)\not\subset A$.
Let's go back to your question: Let $$A=\{(x,0):-1<x<1\}.$$ Your proof that this set $A$ is not closed because (as what you have said) $(-\infty,-1]\cup [1,+\infty)$ is not open is wrong. In here, we are expecting that $A^c\subset\Bbb R^2$.
$(i)$ This is how to show that $A$ is not closed. Observe that 
$$A^c=\{(x,y):y≠0\}\cup\{(x,0):x≤−1\}\cup\{(x,0):x≥1\}.$$
Consider $(1,0)\in A^c$. Then, for any $\epsilon>0$, we have $$B_\epsilon(1,0)\not\subset A^c\qquad\text{do you know why? Try to draw.}$$
This tells us (see Definition 3 applied to the set $A^c$) that the set $A^c$ is not open. Therefore, $A$ is not closed.
$(ii)$ Choose $(x,y)\in A$. Then, for any $\epsilon>0$, we have $$B_\epsilon(x,y)\not\subset A\qquad\text{again, do you know why? Try to draw.}$$
Again, by using Definition 3, we conclude that $A$ is not open.
If you have questions, just ask me. God bless...
