Let $H_1$ and $H_2$ be Hilbert spaces. Is it true that $$B_0(H_1)B_0(H_2, H_1)= \overline{\operatorname{span}}\{ST\mid S \in B_0(H_1), T \in B_0(H_2, H_1)\}= B_0(H_2,H_1)?$$
If so, how can I prove this?
One possible attempt I can think of is to try to show that the left hand side contains all finite-rank operators $H_2 \to H_1$, but I'm not sure these are dense in $B_0(H_2,H_1)?$
Let $F: H_2\to H_1$ be finite rank. Let $P$ denote the orthogonal onto the image of $F$. Since $F$ is finite rank the image of $F$ is finite dimensional (and so closed - hence $P$ is actually well defined) and $P$ is also a finite rank map.
Further $F = P\circ F$ by definition. But this is in $B_0(H_1)\cdot B(H_2,H_1)$. Hence the finite rank maps all lie in this product. Since the finite rank maps are dense in the compacts (we are in the Hilbert space setting) your claim follows.