Let $A$ be a $C^*$-algebra, faithfully represented on the Hilbert space $H$. Thus $A \subseteq B(H)$.
In the literature, I have seen the following $3$ definitions being used to say that $A$ is a non-degenerate $C^*$-subalgebra of $B(H)$:
(1) $[AH]= H$. Here $[AH]= \overline{\operatorname{span}}\{ah: a \in A, h \in H\}$.
(2) $\forall x \in H\setminus \{0\}: \exists u \in A: ux \neq 0.$
(3) $AB_0(H)$ is dense in $B_0(H)$. Here $B_0(H)$ are the compact operators on $H$.
It is easy to see that $(1) \iff (2)$. However, do we also have $(2) \iff (3)$ ? If not, do we at least have an implication or are these concepts not related, although having the same name?
Note that the finite rank maps are dense in the space of compact operators on a Hilbert space. Furhter note that the finite rank maps are spanned by the rank $1$ maps. These can all be written as $v\otimes w^*$, which denotes the map $x\mapsto (x,w)\cdot v$.
Now suppose $(1)$ holds, then for any $v\in H$ there is a sequence $a_n(h_n)\to v$ with $a_n\in A$, $h_n\in H$. Then $$a_n\circ (h_n\otimes w^*) = (a_n(h_n))\otimes w^* \to v\otimes w^*$$ where convergence is in the operator norm. This means that the rank $1$ maps lie in the closure of $AB_0(H)$, hence all of $B_0(H)$ is in the closure.
If we suppose $(3)$ holds we must be able to approximate any compact operator by elements of $AB_0(H)$. In particular for any $x$ lying in the range of a compact operator you must have sequences $a_n,k_n, x_n$ so that $(a_n(k_n(x_n))\to x$. This implies $(1)$, since every element is in the range of some compact operator.