Reproducing Kernel Hilbert Space is dense?

Let's prove the contrapositive.

First, you should check that $R : E^* \to H_\mu$ is the adjoint of $i : H_\mu \to E$. That is, for $h \in H_\mu$ and $x^* \in E^*$, $\langle R x^*, h \rangle = x^*(i(h))$.

Now suppose $i(H_\mu)$ is not dense in $E$. Then by the Hahn-Banach theorem there exists a nonzero $x^* \in E^*$ with $x^*(i(h)) = 0$ for all $h \in H_\mu$. Taking $h = R x^*$, we have that $0 = x^*(i(Rx^*)) = \langle R x^*, R x^* \rangle$. That is, $\int_E |x^*(x)|^2 \mu(dx) = 0$, so as a function on $E$, $x^*$ vanishes $\mu$-a.e. Hence the kernel of $x^*$ is a proper closed subset of $E$ with measure 1, so $\mu$ does not have full support.