Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a uniformly continuous function and $B = \{x \in C : f(x) = 0\}$ its zero set in a closed ball $C$. Assume $dim(B) \leq n-1$. I would like to prove that $B$ can be decomposed into a union of disjoint sets $B = \bigcup_{i=1}^\ell B_i$ with $f(x)=0$ in each $B_i$ and $\ell\geq 1$, such that $\text{dis}(B_i,B_j) = \min\{\|x-y\|,x\in B_i, y\in B_j\} \geq \varepsilon$ for some positive $\varepsilon$ and all $i \neq j$ (or find a counterexample in case this does not hold).
My intuition is that $B$ can be always decomposed into disjoint sets with $\ell\geq 1$, I just want to use the uniform continuity property to prove that $\text{dis}(B_i,B_j) \geq \varepsilon$ (which probably has something to do with the finiteness of $\ell$). My idea so far is the following:
Assume that this is not the case, then there exists a sequence $k_1,k_2,\dots$ such that $\lim_{q\to\infty} \text{dis}(B_{k_q},B_{k_{q+1}}) = 0$, then for all positive $a$ there exist $i,j$ such that $\text{dis}(B_i,B_j) < a$. Then I try to find points $z,h$ close to $B_i,B_j$ where $f \neq 0$ and somehow prove $\|f(z)-f(h)\| \geq s$ for some $s$ which would contradict uniform continuity (similar to how one would prove that $sin(\frac{1}{x}$) is not uniformly continuous). The exact steps are not clear, however.