We break down the proof into steps:

1. 1.

   $\mathcal{A}$ is a lattice. This amounts to verifying various laws on the operations:

   • (idempotency of $\vee$ and $\wedge$): Clearly, $U\wedge U=U$. Also, $U\vee U=(U\cup U)^{{\bot\bot}}=U^{{\bot\bot}}=U$, since $U$ is regular open.

   • Commutativity of the binary operations are obvious.

   • The associativity of $\wedge$ is also obvious. The associativity of $\vee$ goes as follows: $U\vee(V\vee W)=(U\cup(V\cup W)^{{\bot\bot}})^{{\bot\bot}}=U^{\bot}\cap(V\cup W)^{{\bot\bot\bot}}=U^{\bot}\cap(V\cup W)^{\bot}$, since $V\cup W$ is open (which implies that $(V\cup W)^{\bot}$ is regular open). The last expression is equal to $U^{\bot}\cap(V^{\bot}\cap W^{\bot})$. Interchanging the roles of $U$ and $W$, we obtain the equation $W\vee(V\vee U)=W^{\bot}\cap(V^{\bot}\cap U^{\bot})$, which is just $U^{\bot}\cap(V^{\bot}\cap W^{\bot})$, or $U\vee(V\vee W)$. The commutativity of $\vee$ completes the proof of the associativity of $\vee$.

   • Finally, we verify the absorption laws. First, $U\wedge(U\vee V)=U\cap(U\cup V)^{{\bot\bot}}=U^{{\bot\bot}}\cap(U\cup V)^{{\bot\bot}}=(U^{\bot}\cup(U\cup V)^{\bot})^{\bot}=(U^{\bot}\cup(U^{\bot}\cap V^{\bot})^{\bot}=(U^{\bot})^{\bot}=U$. Second, $U\vee(U\wedge V)=(U\cup(U\vee W)^{{\bot\bot}}=U^{{\bot\bot}}=U$.

2. 2.

   $\mathcal{A}$ is complemented. First, it is easy to see that $\varnothing$ and $X$ are the bottom and top elements of $\mathcal{A}$. Furthermore, for any $U\in\mathcal{A}$, $U\wedge U^{{\prime}}=U\cap U^{\bot}=U\cap(X\setminus\overline{U})\subseteq\overline{U}\cap(X\setminus\overline{U})=\varnothing$. Finally, $U\vee U^{{\prime}}=(U\cup U^{\bot})^{{\bot\bot}}=(U^{\bot}\cap U^{{\bot\bot}})^{\bot}=(U^{\bot}\cap U)^{\bot}=\varnothing^{\bot}=X$.

3. 3.

   $\mathcal{A}$ is distributive. This can be easily proved once we show the following: for any open sets $U,V$:

   $$(*)\qquad U^{{\bot\bot}}\cap V^{{\bot\bot}}=(U\cap V)^{{\bot\bot}}.$$

   To begin, note that since $U\cap V\subseteq U$, and ${}^{\bot}$ is order reversing, $(U\cap V)^{{\bot\bot}}\subseteq U^{{\bot\bot}}$ by applying ${}^{\bot}$ twice. Do the same with $V$ and take the intersection, we get one of the inclusions: $(U\cap V)^{{\bot\bot}}\subseteq U^{{\bot\bot}}\cap V^{{\bot\bot}}$. For the other inclusion, we first observe that

   $$U\cap\overline{V}\subseteq\overline{U\cap V}.$$

   If $x\in$ LHS, then $x\in U$ and for any open set $W$ with $x\in W$, we have that $W\cap V\neq\varnothing$. In particular, $U\cap W$ is such an open set (for $x\in U\cap W$), so that $(U\cap W)\cap V\neq\varnothing$, or $W\cap(U\cap V)\neq\varnothing$. Since $W$ is arbitrary, $x\in$ RHS. Now, apply the set complement, we have $(U\cap V)^{\bot}\subseteq U^{\complement}\cup V^{\bot}$. Applying ${}^{\bot}$ next we get $(U\cap V)^{{\bot\bot}}$ for the LHS, and $(U^{\complement}\cup V^{\bot})^{\bot}=U^{{\complement-\complement}}\cap V^{{\bot\bot}}=U^{{\complement\complement}}\cap V^{{\bot\bot}}=U\cap V^{{\bot\bot}}$ for RHS, since $U^{\complement}$ is closed. As ${}^{\bot}$ reverses order, the new inclusion is

   $$(**)\qquad U\cap V^{{\bot\bot}}\subseteq(U\cap V)^{{\bot\bot}}.$$

   From this, a direct calculation shows $U^{{\bot\bot}}\cap V^{{\bot\bot}}\subseteq(U^{{\bot\bot}}\cap V)^{{\bot\bot}}\subseteq(U\cap V)^{{\bot\bot\bot\bot}}=(U\cap V)^{{\bot\bot}}$, noticing that the first and second inclusions use $(**)$ above (and the fact that ${}^{{\bot\bot}}$ preserves order), and the last equation uses the fact that for any open set $W$, $W^{\bot}$ is regular open. This proves the $(*)$.

   Finally, to finish the proof, we only need to show one of two distributive laws, say, $U\wedge(V\vee W)=(U\wedge V)\vee(U\wedge W)$, for the other one follows from the use of the distributive inequalities. This we do be direct computation: $U\wedge(V\vee W)=U\cap(V\cup W)^{{\bot\bot}}=U^{{\bot\bot}}\cap(V\cup W)^{{\bot\bot}}=(U\cap(V\cup W))^{{\bot\bot}}=((U\cap V)\cup(U\cap W))^{{\bot\bot}}=((U\wedge V)\cup(U\wedge W))^{{\bot\bot}}=(U\wedge V)\vee(U\wedge W)$.

Since a complemented distributive lattice is Boolean, the proof is complete. ∎

Clearly, every clopen set is regular open. In addition, $1\in\mathcal{B}$. If $U$ is clopen, so is the complement of its closure, and hence $U^{{\prime}}\in\mathcal{B}$. If $U,V$ are clopen, so is their intersection $U\wedge V$. Similarly, $U\cup V$ is clopen, so that $U\vee V=U\cup V$ is clopen also. ∎

Let $V=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}$. For any $U\in\mathcal{K}$, $U\subseteq\bigcup\{U\mid U\in\mathcal{K}\}$ so that $U=U^{{\bot\bot}}=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}=V$. This shows that $V$ is an upper bound of elements of $\mathcal{K}$. If $W$ is another such upper bound, then $U\subseteq W$, so that $\bigcup\{U\mid U\in\mathcal{K}\}\subseteq W$, whence $V=(\bigcup\{U\mid U\in\mathcal{K}\})^{{\bot\bot}}\subseteq W^{{\bot\bot}}=W$. The infimum is proved similarly. ∎