An open interval is an open set?

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,

$$ c=a - (a - c) \lt a - \epsilon\lt x \lt a + \epsilon \lt a + (d - a) = d \iff x \in (c,d) \implies V_{\epsilon}(a) \subseteq (c, d)$$

Note that $x \in V_{\varepsilon}(a)$ if and only if $x > a-\varepsilon \ge c$ and $x < a+\varepsilon \le d$.

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 $V_\mu(a)\subset V_\delta(p)=(c,d)$.