Q:

Proof verification regarding $\epsilon>0$ being arbitrary. 

Let $f:(\mathbb{R}, \tau_c) \rightarrow (\mathbb{R}, \tau)$ be a function (the topologies don't play a role here),  $z\in\mathbb{R}$ a given point and assume $\forall \epsilon>0, [f^{-1}(z-\epsilon,z+\epsilon)]^c=\{x\in \mathbb{R}:\mid f(x)-z\mid \geq \epsilon \}:=A_{\epsilon}$ is countable .  I want to show that as $\epsilon>0$ is arbitrary then actually all $f^{-1}(\mathbb{R} \setminus  \{ z\})$  is countable.  

Looking at $A_ \epsilon=\{x\in \mathbb{R}:\mid f(x)-z\mid \geq \epsilon \}$ if $x \in A_ \epsilon :f(x)=z$ were possible, then $0\geq \epsilon>0$ which is a contradiction.  Otherwise, if $x \in A_ \epsilon:f(x) \neq z$ then it is possible since there will always have an $ \epsilon>0: \mid  f(x)-z\mid  \geq \epsilon$. I conclude that there are only countable points such that $x \in  f^{-1}(\mathbb{R} \setminus  \{ z\})$.
Do you think it is ok? 
Another aproach would be to set $\epsilon=1/n$ and then show that $\cup_{n=1}^\infty f^{-1}[(-\infty,z-1/n]\cup [1/n+z,\infty)]=f^{-1}(\mathbb{R}\setminus \{z\})$ which is countable since each preimage is countable and the countable union of them is also countable. 
So, as $\epsilon>0$ is arbitrary, in particular, $\{A_{1/n}:n\in\mathbb{N}\}$ are a collection of countable sets whose union is countable.  If $x\in \cup_{\epsilon>0}A_\epsilon$, then $x$ is in some $A_\epsilon$ and $\exists n: 1/n<\epsilon$ and then $\mid f(x)-z\mid\geq \epsilon>1/n$. Hence $x\in A_{1/n}\subseteq \cup_{n=1}^\infty A_{1/n}$. Analogously, if $x \in \cup_{n=1}^\infty A_{1/n}$, then $x\in A_{1/n}$ for some $n$, and $\exists \epsilon>0:1/n>\epsilon$. This leads to $\cup_{n=1}^\infty A_{1/n}=\cup_{\epsilon>0}A_\epsilon$.
Are both approach acceptable? 
Thanks in advance! 

A:

In you first approach you did not consider the possibility there may be uncountable many $\epsilon$. In other words you have shown simply that if $x \in f^{-1}(\mathbb R / {z}), \exists\epsilon > 0, x \in A_\epsilon$ and if for some $\epsilon$ $x \in A_\epsilon, x \in f^{-1}(\mathbb R / {z}) $.  This does not rule out the case that the set is uncountable since there are uncountable many $\epsilon$.
Your second approach is absolutely correct and the best way to go in fact.