This is probably false but can't find a good counterexample.
Let $X$ be a compact space and consider the algebra $C(X)$ of all scalar-valued continuous functions on $X$ endowed with the supremum norm. Let $G$ be an additive subgroup of $C(X)$ that is discrete. (This implies that there is $\delta > 0$ so that $\|f-g\| \geqslant \delta$ for all $f,g \in G$). Let $R$ be the smallest subring of $C(X)$ containing $G$. Must $R$ be discrete too?
No. Let $X$ be the one-point space, so that $C(X) \cong \mathbb{R}$ (or $\mathbb{C}$ works too if you want to use complex scalars).
Now let $G$ be the subgroup $\frac{1}{2}\mathbb{Z} = \{n/2 : n \in \mathbb{Z}\}$. This is discrete. However, the subring of $\mathbb{R}$ generated by $\frac{1}{2}\mathbb{Z}$ is $$\mathbb{Z}[2^{-1}] = \{n/2^k : n \in \mathbb{Z}, k \in \mathbb{N}\},$$ which is not discrete.