Open interval in $R$ is open set
Let $(a,b) = \{ x \in R \mid a < x < b\}$ Let $c \in (a, b)$ and define $\epsilon$-neighborhood of $c$ as $N_\epsilon (c) = \{x \in R \mid \ |x - c| < \epsilon \}$ Take $\epsilon < min\{ b - c, c - a\}$. $x \in N_\epsilon (c) \Longrightarrow |x - c| < \epsilon \Longleftrightarrow c - \epsilon < x < c + \epsilon$ Since $\epsilon < b - c$ and $\epsilon < c - a$, $ a < x < b $ could be rewritten as $a = c - (c - a) < c - \epsilon < x < c + \epsilon < b = c - (b - c) \Longleftrightarrow x \in (a, b) \Longrightarrow N_\epsilon (c) \subseteq (a, b)$