Show that the variety of semilinear Heyting/Brouwerian algebras is not generated by any single finite Heyting/Brouwerian chain where a Brouwerian algebra is a Heyting algebra which does not include a bottom element and semilinear means that it is a subdirect product of chains.
I was hoping if we assume a single finite chain can generate the variety then we would get a contradiction considering an infinite algebra but I haven't been able to make this work.
Assuming that Brouwerian algebras are the same as Heyting algebras but without the $0$ element in the signature (so that, in the infinite case, they might not have a $0$ element at all), and checking the details of the proof of a well-known characterization of congruences on Heyting algebras, I came to the conclusion that the proof doesn't have any reference to the $0$ element, and so the same characterization holds for congruences on Brouwerian algebras.
So, if $\mathbf B$ is a Brouwerian algebra, and $\theta \subseteq B^2$, then $\theta$ is a congruence of $\mathbf B$ iff for some filter $F$ of $\mathbf B$, $$\theta = \{(a,b)\in B^2 : (a\to b) \wedge (b\to a) \in F\}.$$ This allows us to use the same argument for Heyting or Brouwerian algebras.
So let $\mathbf A$ be a Heyting algebra or a Brouwerian algebra.
Then $\mathbf A$ is subdirectly irreducible iff its congruence lattice has precisely one atom $\mu$ (sometimes called the monolith), which is below every congruence except the identity.
Given the above characterization of congruences, this happens precisely if there is a co-atom $u$ in the lattice reduct of that algebra, so that $$\mu = \{u,1\}^2 \cup \{(a,a):a\in A\}.$$
Now, consider a Heyting or Browerian algebra $\mathbf A$ whose lattice reduct is an infinite chain with an element $u$ such that $u \prec 1$.
Then $\mathbf A$ is subdirectly irreducible, so it can't be a subdirect product of finite algebras.