Is the $\{x \in \Bbb R : x \ne 0\}$ set open, close or neither? 
Is the $\{x \in \Bbb R : x \ne 0\}$ set open, close or neither?

The answer is open. However, I can't see why. 
Set $A :=\{x \in \Bbb R : x \ne 0\}$. Following the definition it will be open if for every $a$ from $A$ and any positive $ϵ$ the $V_ϵ(a)$ ⊆ $A$. Now, my intuition says, that if we take $a$ such that it will be really close to $0$ and $ϵ$ be big enough, then $ϵ$ neighbourhood would not be contained in $A$ since it will contain $0$ too.
Can you, please, explain it to me?
 A: You have to be very careful with the quantifiers here.
A set $A$ will be open if for every $a \in A$ there is some $\epsilon$ which has $V_{\epsilon} \subseteq A$, not all $\epsilon$.
A: There is an error in your definition of open set. It doesn't have to hold for any positive $\epsilon$ but for some positive $\epsilon$. Now do you see why $A$ is open?
A: To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$. 
We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.
In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.
A: We know that this set is open since $$\forall x\in A\quad,\quad\exists \epsilon\quad,\quad (x-\epsilon,x+\epsilon)\in A$$for example take $\epsilon=\dfrac{|x|}{10}$ , but it is not closed. Take the convergent sequence $$a_n=\dfrac{1}{n}$$whose limit doesn't belong to $A$ while the terms do.
