The question
Let $E \subseteq \mathbb{R}^n$ and $f \in R(E)$, i.e. ($f\colon E \to \mathbb{R}$). Does the fact that $f$ is continuous on $E$ imply $\mu(E) > 0$?
Reasoning
I mean the fact that $f$ is continuous on $E$ implies (for $a \in E$) that given $\epsilon > 0$ there is some $\delta > 0$ such that $$f( B(a, \delta)) \subseteq B(f(a), \epsilon) \Rightarrow \forall x \in B(a, \delta), f(x) \leq f(a) + \epsilon,$$ but this implies that $$\int_{B(a,\delta)} f(x)dx \leq \int_{B(a,\delta)} (f(a) + \epsilon) dx= (f(a) + \epsilon) \mu(B(a,\delta)),$$ hence in general we cannot have $\mu(B(a,\delta)) = 0$.
Followup
As a response to the given answer,
The part b) implies that since $U_E^\delta(a) \subset E$, we cannot have $\mu(E) = 0$, doesn't it?
Short answer
No, it does not.
Example
Any constant function $f$ will be continuous, no matter the set $E$.
Why does your argument not work?
The definition of continuity talks about metric balls within the metric space on which $f$ is defined. So lets write $B_E$ for the metric balls in $E$, i.e. $B_E(a, \varepsilon) = \{p \in E : d(a, p) < \varepsilon\}$, and $B_R$ for the metric Balls in $\mathbb R^n$, i.e. $B_R(a, \varepsilon) = \{p \in \mathbb R^n : d(a, p) < \varepsilon\}$.
We clearly still have $B_E(a, \varepsilon) = B_R(a, \varepsilon) \cap E \subseteq B_R(a, \varepsilon)$, but we only know $\mu(B_R(a, \varepsilon)) > 0$, whereas $\mu(B_E(a, \varepsilon)) = 0$ is still possible.
Followup
Yes, by $U_E^\delta(a) = B_E(a, \delta) \subseteq E$ we have $0 < \mu(U_E^\delta(a)) \leq \mu(E)$.