A finite quotient and its dual inclusion

The family $I$ contains zero, is downward closed and is closed under union; it omits $\{0,1,2\}$, so is proper. The condition $A\mathrm{△}{A}^{\mathrm{\prime }}\subseteq \{2\}$ is equivalent to $A\cap \{0,1\}=A^{\mathrm{\prime }}\cap \{0,1\}$: membership at $0,1$ must agree, while membership at $2$ is unrestricted. The four classes are $\{\mathrm{\varnothing },\{2\}\}$, $\{\{0\},\{0,2\}\}$, $\{\{1\},\{1,2\}\}$ and $\{\{0,1\},\{0,1,2\}\}$.

Define $r(A)=A\cap \{0,1\}$. It preserves unions and intersections, sends the two bounds to $\mathrm{\varnothing }$ and $\{0,1\}$, and satisfies $r(\{0,1,2\}\setminus A)=\{0,1\}\setminus r(A)$. Its zero fibre is $I$. F2 factors it through $B\mathrm{/}I$; step 1.1 shows the factor sends the four classes bijectively to $\mathrm{\varnothing },\{0\},\{1\},\{0,1\}$. Thus it is the asserted Boolean isomorphism.

The atoms of each powerset algebra are its singletons, so F3 identifies its ultrafilters with point filters. For $i=0,1$, the inverse image of the point filter $V_i$ in $\mathcal{P}(\{0,1\})$ is $r^{-1}[V_i]=\{A:i\in r(A)\}=\{A:i\in A\}=U_i$ in $B$. Its image contains exactly $U_0,U_1$, omitting $U_2$. By F4 the preimage of any basic clopen $[A]$ is the set of those $V_i$ with $i\in A\cap \{0,1\}$, explicitly the clopen $[r(A)]$. This is the stated two-point inclusion with its discrete topology. QED.