Is it true that $\epsilon \rightarrow 0^{+}$ , $(a,b)=[a+\epsilon ,b-\epsilon ]$? The background is:
$\epsilon \rightarrow 0^{+}$    , $f$ is a continuous function on $[a+\epsilon ,b-\epsilon ]$ . Can ''$f$ is continuous in $(a,b)$'' be deduced ?  
I wrote something very strange:
$$(a,b)=(a,a+\epsilon )\cup [a+\epsilon ,b-\epsilon ]\cup (b-\epsilon ,b)$$
prove $(a,a+\epsilon )$ and $(b-\epsilon ,b)$ are $\varnothing$ .
Take $x_{0}\in (a,a+\epsilon )$ , we have:
(1) $x_{0}>a$ .
(2) $x_{0}-a<a+\epsilon -a$ $\Rightarrow$ $x_{0}<a+\epsilon$ $\Rightarrow$ $x_{0}\leq a$
contradiction . So $(a,a+\epsilon )$ and $(b-\epsilon ,b)$ are $\varnothing$ .

Q:(1) Is the background true ?
     (2) Is it true that $\epsilon \rightarrow 0^{+}$  , $(a,b)=[a+\epsilon ,b-\epsilon ]$ ?

 A: No, given $\epsilon>0$, $(a,a+\epsilon )=\emptyset$ is not true. In fact $x_{0}<a+\epsilon$ does not imply that $x_0\leq a$.  
Instead if $a<x_0<b$ there exists $\epsilon>0$ such that $a+\epsilon<x_0<b-\epsilon$ (take for example $\epsilon=\frac{\min(x_0-a,b-x_0)}{2}$). In this way, we show that 
$$(a,b)=\bigcup_{\epsilon>0}(a+\epsilon ,b-\epsilon ).$$
This means that if $x_0\in (a,b)$ then there is $\epsilon>0$ such that $x_0\in (a+\epsilon ,b-\epsilon)$ and therefore if $f$ is continuous in $(a+\epsilon ,b-\epsilon )$ for all $\epsilon >0$ then we may conclude that $f$ is continuous in $(a,b)$
A: For any such $\epsilon$, we have $[a+\epsilon,b-\epsilon]\subsetneq (a,b)$. So no, there is no way to choose an $\epsilon$ that makes $[a+\epsilon,b-\epsilon]=(a,b)$ true. In particular, saying they are equal for all $\epsilon$ is not correct.
However, that's not what you're asked about. You are asked whether something that holds for any $[a+\epsilon,b-\epsilon]$ also holds for $(a,b)$. That is a very different question.
If the function is not continuous on $(a,b)$, then there must be a $c\in(a,b)$ such that $f$ is discontinuous at $c$. And this $c$ is contained in some $[a+\epsilon,b-\epsilon]$.
This proves the contrapositive of the statement you want to prove, so it proves your statement.
A: You need to be careful in applying the "for all" quantifier.  The logical statement 
$$\forall\epsilon\gt0\,\,\,(x_0\lt a+\epsilon\implies x_0\le a)$$ 
does not hold. What does hold is the logical statement 
$$(\forall\epsilon\gt0\,\,\, x_0\lt a+\epsilon)\implies x_0\le a$$
In other words, it's not true that $(a,a+\epsilon)=(b-\epsilon,b)=\emptyset$, but, rather
$$\bigcap_{\epsilon\gt0}(a,a+\epsilon)=\bigcap_{\epsilon\gt0}(b-\epsilon,b)=\emptyset$$
