Let $H$ be an infinite dimensional separable Hilbert space.
Let $B(H)$ be the algebra of bounded operators on $H$.

Definition : $T \in B(H)$ is quasi-diagonalizable if $\exists r \in \mathbb{N}$ and $(e_{n})_{n \in \mathbb{N}}$ orthonormal basis :
$$ \forall n , m \in \mathbb{N} \ , \ (T.e_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r $$

Invariant subspace problem (ISP): Is it true that for all $T \in B(H)$ there exists a proper, closed and $T$-invariant subspace $K \subset H$ (i.e. $T(K) \subset K = \overline{K} \subsetneq H$)?

Remark: This problem is known for the Banach spaces in general, but still open for an Hilbert space.

Is it true that the quasi-diagonalizable operators check the ISP?

Remark: I post the following question on MSE : Are all the operators quasi-diagonalizable?
It's probably false, but of course, if it's true, the question here would be equivalent to the ISP, and I would be interested to know if this way is (in your opinion) promising, and ever known (reference).