Verification : Prove every variety is generated by the class of subdirectly irreducible algebras in the class. Prove every variety is generated by the class of subdirectly irreducible algebras in the class.
Let $\mathcal{V(K)}$ be a variety for the class $\mathcal{K}$ of similar type algebras and let $\mathcal{K'}$ be the class made of all the subdirectly irreducible algebras in $\mathcal{K}$. Now, every algebra $A$ is isomorphic to a subdirect product of subdirectly irreducible algebras. AS $\mathcal{V(K)}$ is a variety we get $\forall A \in \mathcal{V(K)}$, $A $ is isomorphic to a subdirect product of subdirectly irreducible algebras $B_{i \in I} \in \mathcal{V(K)}$. Hence by taking all possible subdirect products of members of $\mathcal{K'}$. We are able to generate isomorphic copies of each $A \in \mathcal{V(K)}$. So we get $\mathcal{V(K)} \le \mathcal{V(K')}$ so we need to make sure we do not generate any more than that. However, we do not as $\mathcal{K'} \subseteq \mathcal{K}$ thus $\mathcal{V(K')} \subseteq \mathcal{V(K)}$. Hence, $\mathcal{V(K')} = \mathcal{V(K)}$.
 A: Your approach to the problem is a little convoluted. I assume that you are trying to prove that 

every variety $\mathcal{W}$ is generated by the subdirectly irreducible algebras in $\mathcal{W}$, 

but at a first glance you seem to try proving the (false) statement every variety generated by a class $\mathcal{K}$ is generated by the subdirectly irreducible algebras in the class. To see this is false, take the two-element lattice $\boldsymbol{2}$ and some non-distributive lattice $\mathbf{L}$ and $\mathcal{K}:=\{\boldsymbol{2},\mathbf{L}\times\mathbf{L}\}$ as your class.
Hint to prove the result: if $\mathbf{A}$ is the subdirect product of $\mathbf{A}_i$, then $\mathbf{A}_i\in H(\{\mathbf{A}\})$ for all $i$.
Hover for the rest  of the proof.

 Call $\mathcal{W}_{SI}$ the class of subdirectly irreducibles of $\mathcal{W}$. Let $\mathbf{A}\in\mathcal{W}$. Then since $\mathcal{W}$ is a variety, it is closed under taking homomorphic images. Hence every subdirect factor of $\mathbf{A}$ is a member of $\mathcal{W}_{SI}$. Hence every $\mathbf{A}\in\mathcal{W}$ is a subalgebra of a product of elements of $\mathcal{W}_{SI}$, and a fortiori belongs to the variety generated by it.

