Let $B$ be a $C^*$-subalgebra of $A$. Do we have $AB= A?$ In other words, is $\text{span}\{ab \mid a \in A, b \in B\}$ norm-dense in $A$?
The usual way to approach these questions is by using approximate units, so that's how my attempt will go. Consider the canonical approximate unit of $A$, that is $$(u_i)_{i \in I}$$ where $I$ is the set of all positive elements in the closed unit ball of $A$ and $u_i= i$.
Then if $a \in A$, we have $$a = \lim_i au_i$$
Now, consider the subnet $(u_j)_{j \in J}$ where $J$ is the set of all positive elements in the closed unit ball of $B$. Then we still have $$a= \lim_j a u_j \in AB$$
Is this correct?
This is not true. Consider $A=M_2(\mathbb C)$, and $$B=\left\{\begin{bmatrix}b&0\\0&0\end{bmatrix}:b\in\mathbb C\right\}.$$ Note that $$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b&0\\0&0\end{bmatrix}=\begin{bmatrix}a_{11}b&0\\a_{21}b&0\end{bmatrix},$$ so clearly $AB$ cannot be all of $A$.