A question related to interior points 1. Every point of the interval $(0, 1)$ is an interior point of that interval. Thus $(0, 1)^o = (0, 1)$.
2. Let $A = [a, b]$ where $a < b$. Then $A^o = (a, b)$.
I am trying to prove the $2^\text{nd}$ statement. The two problems above look almost identical. The difference is that an arbitrary element $x$ is assumed to be in  $(0, 1)$ in the $1^\text{st}$ one and $x$ is assumed to be in  $[a, b]$ in the $2^\text{nd}$. Does that make any difference? 
Thanks. 
 A: If you want to use the first result to prove the second. Just realize that you can 'scale' the interval $(0,1)$ to become any interval $(a,b) \subset \mathbb{R}$. But you are looking at $[a,b]$. So show that neither $a$ nor $b$ can be interior points for $[a,b]$. Then all that is left of the interval is $(a,b)$, which so long as you give a convincing argument to as why this 'scaling' does not change the interior of an interval, you are done. 
However, you don't have to do it this way since you can just as easily (if not more so) prove directly that the interior of $[a,b[$ is $(a,b)$ in $\mathbb{R}$. Just as yourself, given any point $c$ of $[a,b]$, can you find an open interval $(s,t)$ containing $c$ that is contained in $[a,b]$? Which is there no such interval for $a$ and $b$ but is for every $c \in [a,b]$ so long as $c \neq a$ and $c \neq b$? Knowing that, it is clear what the interior of $[a,b]$ is.
A: Manuel's suggestion is right on the money: for any $\;\epsilon>0\;$ we have that
$$(a-\epsilon\,,\,a+\epsilon)\rlap{\;\,/}\subset [a,b]\implies a\;\;\text{is not an interior point}$$
and something similar for $\;b\;$ , so what you know about $\;(a,b)\;$ applies here...
