A question on open sets If we define open as: A set $O⊆R$ is open if for all points $a∈O$ there exists an
$\epsilon$-neighborhood $V_\epsilon(a)⊆O$.
Where $V_\epsilon(a) = \{x \in \mathbb{R}: | x - a | < \epsilon\}$
Now consider some open interval: 
$(c,d) = \{x \in \mathbb{R} : c<x<d \}$
To see that $(c,d)$ is open, let $x \in (c,d)$ be arbitrary.
let $\epsilon = \text{min}\{x-c,d-x\}$, then it follows that $V_\epsilon(x) \subseteq (c,d)$
I am unable to see why this definition does not hold for a set containing one or more closed end points.
If my understanding is correct, let's take the closed interval $[1,10]$ and lets choose $x = 10$. So clearly $x \in  [1,10]$
$\epsilon = \text{min} \{10-1, 10-10\} = 0$
Then isn't it still true that
$V_\epsilon(x) \subseteq [1,10]$ 
Also, how would you express $V_\epsilon(x)$ in this instance, like it is expressed in the third line?
 A: If you allowed $\epsilon$ to be $0$, then every set is open since the $0$-neighborhood of a point is just itself. For this reason, $\epsilon$ is understood to be positive.
A: You correctly focused on one of the end points.  For every point that is not an end point in the interval $[1,10]$, your argument for finding a ball around that point contained in $[1,10]$ works.
Now, if $x = 10$, we need to show that every ball around $10$ is not contained in $[1, 10]$.  Ok, so let $\epsilon > 0$ be arbitrary.  $V_{\epsilon}(10) = (10 - \epsilon, 10 + \epsilon)$.  This clearly contains $10$ for each positive $\epsilon$.  BUT, is this interval contained in $[1,10]$?  Of course not.  The largest number in $[1,10]$ is $10$, but for each positive $\epsilon$, $(10 - \epsilon, 10 + \epsilon)$ contains a number bigger than $10$, and so this open set is not contained in $[1,10]$.
So, for every positive $\epsilon$, the ball around $10$ of radius $\epsilon$, which is equal to the interval $(10 - \epsilon, 10 + \epsilon)$, is not contained in $[1,10]$, which means $[1,10]$ is not open.
Instead of $x = 10$, you could have made a similar argument for $x = 1$.  I will leave it to you as an exercise to show that $x = 1$ causes the definition of open to fail for the set $[1,10]$.
