We want to show that the set $S=\{x\in\mathbb R : 1<x<3\}$ is open using only the definition of open sets.
[ Definition: A subset $S$ of $\mathbb R^n$ is open iff $\forall s\in S, \exists B_{\epsilon}(s)=\{x\in\mathbb R^n : ||x-s||<\epsilon\}$ such that $B_{\epsilon}(s)\subset S$ ]
Proof:
Start with an arbitrary $s\in S=(1,3)$ and an arbitrary $\epsilon>0$. Construct an $\epsilon$-ball $\{x\in\mathbb R : |x-s|<\epsilon\}$. Now, we have
$$|x-s|<\epsilon \iff -\epsilon<x-s<\epsilon \iff s-\epsilon<x<s+\epsilon$$
or $x\in(s-\epsilon,s+\epsilon)$. I am not sure how to conclude this proof. Essentially, how to conclude that $B_{\epsilon}\subset S$.
By definition, for every $ s\in S$, there exist $\varepsilon>0$, such that $B_\varepsilon(s) \subset S$. Just take $$\varepsilon=\min\{3-s,s-1\}$$ Then this $\varepsilon $ works!
Verification: Let $x \in B_\varepsilon(s)$.Then $|x-s|<\varepsilon.$ That is, $$x \in (s-\varepsilon,s+\varepsilon)$$.
If $s \geq 2$, then $\varepsilon=3-s.$In this case $$x \in (s-\varepsilon,s+\varepsilon)=(2s-3,3) \subset(1,3)$$
If $s \leq 2$, then $\varepsilon=s-1.$In this case also $$x \in (s-\varepsilon,s+\varepsilon)=(1,2s-1) \subset(1,3)$$