In the book "Understanding Analysis" by Stephen Abbott, the author defines an open set as:
A set $O \subseteq \mathbb{R}$ is open if for all points $a \in O$ there exists an $\epsilon$-neightborhood $V_\epsilon (a) \subseteq O$.
Here the $\epsilon$-neighborhood is defined as: $$ V_\epsilon(a) = \{ x \in \mathbb{R} \mid |x-a| < \epsilon \} $$
Now, according to the above definitions, if we consider the closed interval $[c,d]$, then it appears to me that $[c,d]$ is not an open set (because for the points $c$ and $d$ we cannot find an $\epsilon$-neighborhood such that $V_\epsilon(c) \subseteq [c,d]$ nor $V_\epsilon(d) \subseteq [c,d]$). However, I know that for any metric space $X$ (and $[c,d]$ is a metric space), the whole set $X$ is open in $X$. Thus, something has gone wrong in the above reasoning. Can anybody shed some light on this?
You changed your metric space from $\mathbb{R}$ to $X$. So your definition of $V_\epsilon(a)$ needs to change, too:
$$V_\epsilon(a) = \{ x \in X \mid |x-a| < \epsilon \}$$