According to comments by Hamza I revise the question.
Let $H$ be an infinite dimensional separable Hilbert space.
Is there an increasing sequence of subvector spaces $V_{1} \subsetneq V_{2} \subsetneq \ldots \subsetneq V_{n} \ldots $ of $B(H)$ such that each $V_{n}$ is a closed subspace and is closed under $*-$operation.(Say operator system) Moreover we have $$ B(H)V_{n} \subset V_{n+1}$$
that is $ax \in V_{n+1}$ if $a\in B(H),\;\;x\in V_{n}$
Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators.
If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count carefully, so this might be off by one or two; the point is that every multiplication by a new element pushes you up one step on subspace in the chain), so $V_8=B(H)$.
These ideas can probably be used to argue that the chain has to be even shorter; I'm afraid I don't have the time now.