Show that the interior of a set $S^0$ is open. $S^0$  is the interior of a set $S$.
Let $x\in S^0$ be given. We want to find $\delta>0$ such that $(x-\delta,x+\delta)\subset S^0$.
$S^0$ is the interior of $S$, then $S^0\subset S$, then $x\in S$ and $\exists \delta_1>0$ such that $(x-\delta_1,x+\delta_1)\subset S$.
Can we say choose $\delta<\delta_1$  such that
$(x-\delta,x+\delta)\subset(x-\delta_1,x+\delta_1)$,
then $(x-\delta,x+\delta)\subset S^0$,
therefore the interior of $S$ is open?
Does it make sense? Please guide me.
 A: That makes perfect sense, but I can see that you might be slightly confused yourself.  I'll try and explain your own reasoning!
For $x$ to be in the interior of $X$, you need to show the existence of a $\delta$ such that $(x-\delta,x+\delta)\subseteq X$.  A set is open if it is its own interior, and you are trying to show that the interior of any set is always open.  
So, what you have done is to show that every element of $S^{\circ}$ is an element of $\left(S^{\circ}\right)^{\circ}$.  You have taken an arbitrary element $x\in S^{\circ}$. Since it is in $S^{\circ}$, you know that there is a $\delta$ such that $(x-\delta,x+\delta)\subseteq S$.  So far so good.  But what we need to do is show that $x\in \left(S^{\circ}\right)^{\circ}$.  You have the right elements for this, but I'm not sure you quite know how to put them together.
You have done the right thing by taking $\delta_{1}\leq \delta$.  For simplicity, let's set $\delta_{1}=\delta/{2}$.  Then it will suffice to show that $(x-\delta_{1},x+\delta_{1})\subseteq S^{\circ}$.  So, take an arbitrary $y\in (x-\delta_{1},x+\delta_{1})$.  It follows that 
$$(y-\delta_{1},y+\delta_{1}) \subseteq (x-2\delta_{1},x+2\delta_{1}) = (x-\delta,x+\delta) \subseteq S.$$
But this is sufficient to show that $y\in S^{\circ}$.  There is therefore an open interval about $x$ which is contained in $S^{\circ}$, and so $x\in\left(S^{\circ}\right)^{\circ}$.  Since $x$ is an arbitrary element of $S$, we get 
$$S^{\circ}=\left(S^{\circ}\right)^{\circ}.$$
A: Treating the open interval $(x - \varepsilon, x + \varepsilon)$ as an open ball in the metric sense then $B(\varepsilon, x) := (x - \varepsilon, x + \varepsilon)$. So let x be any point in $Int(S)$, then there exists $r > 0$ such that $B(r,x)$ is  in $S$. Claim: $B(r/2, x)$ is contained in the $Int(S)$. Let y be any point in $B(r/2, x)$, then $B(r/2,y)$ is in $B(r, x)$. To see this, let $z$ be any point in $B(r/2, y)$, then $d(z,y) < r/2$ and $d(y,x) < r/2$. So $d(x,z) < d(x,y) + d(y,z) < r/2 + r/2 = r$. So $z$ belongs to $B(r, x)$. So 
$B(r/2, y)$ is in $B(r, x)$ and $B(r, x)$ is in $S$. So $B(r/2, y)$ is in $S$. This means $B(r/2, x)$ is in $Int(S)$. So $Int(S)$ is open. 
