Showing that an open interval is an open set Context:
I'm trying to algebraically prove that an open interval is an open set. If I sketch it, as suggested by @rschwieb in this answer, then it seems quite obvious that this is indeed true. But I would like to be able to show it algebraically and after having looked at various sources found on the internet, I have decided to ask it here.
Attempt:
For definiteness, let us consider:
\begin{equation}
(c,d) = \{ x \in \mathbb{R} \mid c < x < d \}
\end{equation}
Furthermore, let $a \in (c,d)$, and recall that the $\epsilon$-neighborhood of $a$ is the set:
\begin{equation}
V_\epsilon (a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \} 
\end{equation}
Now, if we take $\epsilon = \mathrm{min} \{ a-c,d-a \}$, then $a-\epsilon \geq c$ and $a+\epsilon \leq d$. Up until here I understand everything. But then I don't understand how we can conclude that $V_\epsilon (a) \subseteq (c,d)$?
 A: Note that $x \in V_{\varepsilon}(a)$ if and only if $x > a-\varepsilon \ge c$ and $x < a+\varepsilon \le d$.
A: Here's an alternate way to go about proving that $V_\epsilon(a) \subset (c,d)$ for some $\epsilon >0$: instead of "measuring distance to the edges," you can "measure distance to the center." Let $p$ be the midpoint of $(c,d)$. Then
$$(c,d) = V_\delta(p),$$
where $\delta:= |p-c|=|p-d|.$ Given $a \in (c,d)$, let $\mu = \delta-|a-p|.$ Given
$x \in V_\mu(a)$: $$|x-p|=|x-a+a-p| \leq |x-a|+|a-p| \leq \delta-|a-p|+|a-p|=\delta.$$
Hence $V_\mu(a)\subset V_\delta(p)=(c,d)$.
A: Choose $\epsilon \lt \mathrm{min} \{ a-c,d-a \}$.
Picture it geometrically by drawing a real line. $ |x - a| \lt \epsilon $ represents all points on the line that are $\epsilon$-distant from the point $a$. By picking $\epsilon \lt \mathrm{min} \{ a-c,d-a \}$ what you do is to pick the smallest distance from the point $c$ to the boundaries of the interval. Now if we create a neighbourhood (an open set) around $c$ again using this minimum distance it will clearly be contained in the original interval. 
Rigorously, 
$x \in V_{\epsilon}(a) \implies |x - a| \lt \epsilon \iff a - \epsilon\lt x \lt a + \epsilon \tag{1}$
Now $\epsilon \le a-c $ and $\epsilon \le d - a$. Use these to approximate $\epsilon$ in $(1)$. That is, 
$$ c=a - (a - c) \lt a - \epsilon\lt x \lt a + \epsilon \lt a + (d - a) = d \iff x \in (c,d) \implies V_{\epsilon}(a) \subseteq (c, d)$$
A: My attempt of a complete answer:
Let $( c,d ) = \left\{ x \in \mathbb{R} : c < x < d \right\} \subseteq \mathbb{R}$ be an open interval and let $a \in ( c,d ) $ be any point from it. Notice that $a-c>0$ and $d-a>0$ are distances from $a$ to the interval's endpoints $c \notin ( c,d ) $ and $d \notin ( c,d )$. Let $\epsilon = \min( a-c, d-a )$. Then, $a - \epsilon \geq c$ and $a + \epsilon \leq d$. A point with distance $\epsilon$ from $a$ isn't contained in the interval (it's the interval's boundary point). Yet, if we create a neighborhood (open interval) $V_{\epsilon}( a )$ containing points for which $|x - a| < \epsilon \iff a - \epsilon < x < a + \epsilon$, we only require the distance from $a$ to be smaller than $\epsilon$, which means that the endpoints $c, d$ are never included in the neighboorhod and hence $V_{\epsilon}( a )$ is contained in $( c,d )$:
$$\forall a \in ( c,d ) \; {and} \; \epsilon=\min( a-c,d-a ),$$
$$x \in V_{\epsilon}( a ) \iff c \leq a - \epsilon < x < a + \epsilon \leq d$$
$$x \in V_{\epsilon}( a ) \iff c < x < d$$
$$x \in V_{\epsilon}( a ) \iff x \in ( c,d )$$ 
    Hence
$$\forall x \in V_{\epsilon}( a ), x \in ( c,d ) \implies V_{\epsilon}( a ) \subseteq ( c,d ) .$$ 
Remark: To me it seems that the main reason for an open interval to be an open set is the strict inequality in the definition of $V_{\epsilon}( a )$, which requires that there must be an $\epsilon>0$ distance to the neighbor. The proof would fail if a point $a$ could be a neighboring point for itself.
A: Yes, an open interval in $\mathbb{R}$ is an open set in the topology generated by the standard metric.
A: For any point $x$ in an open interval $(a,b)$,
let $p=|x-a|$ and $q = |b-x|,$
then $x \in (a+\frac{p}{2}, b-\frac{q}{2}).$
Since $(a+\frac{p}{2}, b-\frac{q}{2})$ is entirely contained in $(a, b)$, then $(a,b)$ is an open set
