The set of all exterior points is an open set 
Let $S$ be a subset of $R$. Then the set of all exterior points of $S$ is an open set.

My proof is as follows:

For any element $x$ in $\operatorname{ext}(S)$ (the set of all exterior points of $S$), $B(x;r)$ for $r>0$ is not a subset of $S$. We know that since $B(x;r)$ is open set, any point in $B(x;r)$ is exterior point of $S$ and thus there is $h>0$ such that $B(x_1;h)$ is not a subset of $S$. $B(x_1;h)$ is a subset of $B(x;r)$, implying $\operatorname{ext}(S)$ is open set.

Is this proof valid? If not can someone help me modify this more properly? 
 A: The exterior is the interior of the complement of the set.
We need to show that every point of the exterior is contained in a ball that consists entirely of points in the exterior.
Recall that interior is the set of all point that can be covered by a ball completely contained in the set.
So, if $x\in\text{ext}(S)$ then there is a ball $B(x,r)$ such that $B(x,r)$ is contained in the complement of $S$, i.e. it doesn't intersect $S$ (notice how doesn't intersect is not the same as is not a subset). Each point $y\in B(x,r)$ is contained in a ball $B(y,r-d(x,y))$ that is contained in $B(x,r)$, and therefore also contained in the complement of $S$. Therefore $y$ is also in the interior of the complement of $S$, i.e. $y$ is also in $\text{ext}(S)$. 
Therefore $\text{ext}(S)$ is open.
A: An easy way :
suppose $x\in Ext(S)$
then $\exists r_x>0 $ such that $B(x,r_x)\subseteq X\setminus S$
Now it remains to show only that $B(x,r_x)\subseteq Ext(S)$
let $y\in B(x,r_x)$ and since $B(x,r_x)$ is open then $\exists s_y>0$ such that $B(y,r_y)\subset B(x,r_x)\subseteq X\setminus S$
thus we have for any $y\in B(x,r_x)$ we have $B(y,r_y)\subseteq X\setminus S$
thus $y\in Ext(S)$.Hence $B(x,r_x)\subseteq Ext(S)$
A: x is in the exterior iff it has neighborhood B(x) disjoint from x iff  it has B(x) subset of the exterior.  The exterior is U {B(x)| x is in the exterior} is the union of open nghbrhoods, is open. 
