Suppose that $\pi$ is a $\ast$-representation of a $C^{\ast}$-algebra $A$ on some Hilbert space $H$. We say that $\pi$ is non-degenerate if for each $0\neq y\in H$ there is an $a\in A$ such that $\pi(a)y\neq0$.
Now suppose that $D\subset H$ is a dense subspace and that for each $0\neq x\in D$ there is an $a\in A$ such that $\pi(a)x\neq0$. (I guess one could say that $\pi$ is non-degenerate on $D$.)
Can we conclude from this that $\pi$ is non-degenerate?
My attempt: Since I couldn't find any counterexamples, I tried to find a proof instead. Let $0\neq y\in H$ be given. Fix some sufficiently small $\varepsilon>0$, which can later be chosen suitably. By density, there exists some $x\in D$ such that $\|x-y\|<\varepsilon$. Since we may assume that $\varepsilon<\|y\|/2$, it follows that $x\neq0$. Hence by assumption there is an $a\in A$ such that $\pi(a)x\neq0$. Now I was hoping that $\pi(a)y\neq0$, but since our choice of $\varepsilon$ may not depend on $a$ and $x$, I don't know how to continue from here.
Nice question. It misguided my intuition at first. The answer turns out to be negative. While trying to prove that the answer was positive, I noted that the main obstruction was that a dense subspace of a Hilbert space must not intersect all closed subspaces. Starting from there, it is not too hard to construct a counterexample.
Choose a Hilbert space $H$ with a dense subspace $D$ and a closed proper subspace $0 \ne M$ such that $M\cap D = 0$ (see Is intersection of a dense subspace and a closed subspace of a Hilbert space also Dense?).
Then the canonical faithful representation $\pi: B(M^\perp) \hookrightarrow B(H): x \mapsto 0 \oplus x$ where $$(0 \oplus x)(m+m^\perp) = xm^\perp \in M^\perp$$ is degenerate (it is non-unital). However, given $0 \ne d \in D\subseteq M \oplus M^\perp$, write $d = m + m^\perp$. Note that $m^\perp \ne 0$, since $M \cap D=0$. But then we can choose $x \in B(M^\perp)$ with $x m^\perp \ne 0$ and then $$\pi(x)d = x m^\perp \ne 0.$$