The following seems to be a well-known theorem in operator algebras:
$\textbf{Theorem}.$ The only irreducible $C^*$-subalgebra of the compact operators $K(\mathcal{H})$ is itself. That is, for a $C^*$-subalgebra $\mathcal{A}$ of $K(\mathcal{H})$, if $\operatorname{id}:\mathcal{A} \to B(\mathcal{H})$ is irreducible, then $\mathcal{A}=K(\mathcal{H}).$
Does it immediately follow from this theorem that there exists no (nonzero) subrepresentations of $\operatorname{id}:K(\mathcal{H})\to B(\mathcal{H})$ aside from itself?
This seems pretty immediate to me (unless I'm misunderstanding something) because for a subrepresentation of $\operatorname{id}:K(\mathcal{H})\to B(\mathcal{H})$ to exist, there needs to be a closed subspace of $\mathcal{H}$ which is inviariant in $\operatorname{id},$ which cannot happen unless this subspace is $\{0\}$ of $\mathcal{H}$ according to the theorem above.
A related question: Does this mean that if $\mathcal{J}$ is an ideal of $K(\mathcal{H})$ then for any $h\in \mathcal{H},$ $[\mathcal{J}h]=\mathcal{H}$ (provided $\mathcal{J}$ is non-trivial)? Because $[\mathcal{J}h]$ is invariant with respect to $\operatorname{id:K(\mathcal{H})\to B(\mathcal{H})}.$
You don't need the theorem for this. The C$^*$-algebra $K(\mathcal H)$ is simple. And $K(\mathcal H)$ is irreducible, since $K(\mathcal H)''=B(\mathcal H)$, which means that $K(\mathcal H)$ has no reducing subspaces (equivalently, no projections in its commutant).