First few definitions:
$A \in I(K)$ iff $A$ is isomorphic to some member of $K$
$A \in S(K)$ iff $A$ is a subalgebra of some member of $K$
$A \in H(K)$ iff $A$ is a homomorphic image of some member of $K$
$A \in P(K)$ iff $A$ is a direct product of a nonempty family of algebras in $K$
$A \in P_s(K)$ iff $A$ is a subdirect product of a nonempty family of algebras $in$ K.
I have proved that $SH \leq HS$ I am able to do this. But I need to prove that $HS \neq SH$.
The only homomorphic images of a field $F$ (when it is considered as a ring) are the field itself and the zero ring $\{0\}$.
If we consider the ring $(\mathbb{Q},+,\cdot,-,0,1)$ we see that $\mathsf{SH}(\mathbb{Q})$ will just contain $\{0\}$ and all the subrings of $\mathbb{Q}$. Note that every subring of $\mathbb{Q}$ will have characteristic 0.
If we look at $\mathsf{HS}(\mathbb{Q})$, we can see that $\mathbb{Z}$ is a subring of $\mathbb{Q}$ and $\mathbb{Z}/p\mathbb{Z}$ is a homomorphic image of $\mathbb{Z}$. But $\mathbb{Z}/p\mathbb{Z}$ has characteristic $p$, so it is not in $\mathsf{SH}(\mathbb{Q})$. Thus $\mathsf{SH}(\mathbb{Q})\neq \mathsf{HS}(\mathbb{Q})$.