If S $\subseteq \Bbb R$ and $x \in \text{int}(S)$, then $\exists r>0$ such that the open ball $B_r(x) \subseteq S$. It is known that the open balls are open sets in $\Bbb R$ then $B_r(x)$ is open. If $y \in B_r(x)$, then there is some $r_0>0: B_{r_0} \subseteq B_r(x)$, so $y \in \text{int}(S)$ because
$$\text{int}(S) := \{z\in S\mid\exists \varepsilon > 0: B_{\varepsilon}(z) \subseteq S\} $$
which is equivalent to
$$\text{int}(S) = \bigcup_{x\in \text{int}(S)} B_{\varepsilon_x}(x)$$
then $B_{r_0}(y) \subseteq \text{int}(S)$, therefore $\text{int}(S)$ open and because $S$ is arbitrary, the interior of $any$ set is open.
I'm not 100% sure if the proof is right, particularly the part that i say that the interior of $S$ is the union of the open balls of its elements.
Best Answer
Take any $x \in Int \ S$. I will show you that there exists a ball $N_x$ around $x$ completely contained in $Int \ S $.
By assumption, there exists a ball $N_x$ with $N_x \subseteq S $. So take any point $y \in N_x$. As you note, balls are open, so there exists a ball $N_y \subseteq N_x \subseteq S$ around $y$. But this means by definition that $y \in Int \ S$. Thus, if $y\in N_x$, then $y\in Int \ S$, and so $N_x \subseteq Int \ S $, and $Int \ S$ is open.