Let $\mathbb{A}=(A,\mathcal{F}^\mathbb{A})$ be a countable algebra. I need to prove that $|Sub(\mathbb{A})|\leqslant\aleph_0$ or $|Sub(\mathbb{A})|=2^{\aleph_0},$ where $Sub(\mathbb{A})$ is the lattice of subuniversums of $\mathbb{A}.$
I suppose I have to use topology and Cantor set.
Give the set $2$ the discrete topology, and then give $2^{\omega}$ the product topology. The resulting space is (or is homeomorphic to, depending to your definition) the Cantor space.
If $A$ is an algebra with universe $\omega$, then any subalgebra $S\leq A$ is a subset of $A$, hence has a characteristic function $\chi_S\in 2^{\omega}$. The collection $X$ of these characteristic functions is in bijective correspondence with Sub($A$), so it suffices to prove that the set $X$ of characteristic functions is countable or of cardinality $2^{\aleph_0}$.
$X$ is a closed subset of $2^{\omega}$, since subalgebra generation is an algebraic closure operator. By the Cantor–Bendixson theorem, a closed subset of the Cantor space is the union of a countable set and a perfect set. So, if $X$ is uncountable, then it contains a perfect subset, from which one gets $|X|\geq 2^{\aleph_0}$. (Perfect subsets of the Cantor set have size continuum.) The reverse inequality follows from $X\subseteq 2^{\omega}$.