Closure and complement generate at most fourteen sets from any subset, and $(0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$ attains fourteen

$cc = 1$ and $kk = k$ hold by [A3].

For an open $B \subseteq X$: $B \subseteq kB$ by [A3], and $B = iB$ because $B$ is open, so monotonicity of $i$ gives $B = iB \subseteq ikB$ and monotonicity of $k$ gives $kB \subseteq kikB$; conversely $ikB \subseteq kB$ gives $kikB \subseteq kkB = kB$. Hence $kikB = kB$ for every open $B$.

In $\mathbb{R}$, for $a < b$ the interval $(a,b)$ is open: for $x \in (a,b)$ put $r := \min\{x - a,\ b - x\} > 0$ by [L3]; then $(x-r, x+r) \subseteq (a,b)$. The rays $(-\infty,b)$ and $(a,\infty)$ are open by the same computation with one of the two bounds omitted.

In $\mathbb{R}$, for $a < b$ and any $x \in [a,b]$ and any $r > 0$, the interval $J := (\max\{a,\ x-r\},\ \min\{b,\ x+r\})$ is nonempty: its left endpoint is below its right endpoint because $a < b$, $a \le x < x + r$ and $x - r < x \le b$ and $x - r < x + r$. Every point of $J$ lies in $[a,b] \cap (x-r, x+r)$.

No nonempty open subset of $\mathbb{R}$ is contained in $\mathbb{Q}$: it would contain a ball $(x-r,x+r)$, which is uncountable by [L2], whereas a subset of $\mathbb{Q}$ is at most countable by [L2].

In $\mathbb{R}$, for $a \le b$ the interval $[a,b]$ is closed, its complement being $(-\infty,a) \cup (b,\infty)$, a union of two open sets; likewise $(-\infty,b]$ and $[a,\infty)$ are closed, and a singleton $\{t\}$ is closed, its complement being $(-\infty,t) \cup (t,\infty)$.

By step 1.4 and [L2] the set $J$ contains a rational, and being a nonempty open interval it is uncountable by [L2] while $\mathbb{Q}$ is at most countable, so $J$ also contains a point outside $\mathbb{Q}$. Hence for every $x \in [a,b]$ and every $r > 0$ the ball $(x-r,x+r)$ meets both $[a,b] \cap \mathbb{Q}$ and $[a,b] \setminus \mathbb{Q}$.

Claim 1: by step 1.2 applied to the open set $B := iA$ one gets $kikiA = kiA$ for every $A$, that is $kiki = ki$ as operators; substituting $i = ckc$ turns $ki$ into $kckc$ and gives $kckckckc = kckc$; composing on the right with $c$ and using $cc = 1$ gives $kckckck = kck$.

Consequently $\overline{[a,b] \cap \mathbb{Q}} = [a,b] = \overline{[a,b] \setminus \mathbb{Q}}$ for $a < b$: each closure is contained in $[a,b]$, which is closed by step 2.1, and contains $[a,b]$ by step 2.2 together with the neighbourhood criterion for the closure.

Likewise $\overline{(a,b)} = [a,b]$ for $a < b$: the inclusion $\subseteq$ holds because $[a,b]$ is closed and contains $(a,b)$, and $\supseteq$ because for $x \in [a,b]$ and $r > 0$ the nonempty interval $J$ of step 1.4 meets $(a,b)$, being contained in $(a,b)$ except possibly for its endpoints, which it excludes. The same argument gives $\overline{(a,\infty)} = [a,\infty)$ and $\overline{(-\infty,b)} = (-\infty,b]$.

Claim 2: using $cc = 1$ and $kk = k$, every word in $k$ and $c$ equals an alternating word, one with no two adjacent equal letters. An alternating word of length at least $8$ contains $kckckck$ as a block of seven consecutive letters — the first seven if it begins with $k$, the second through eighth if it begins with $c$ — and replacing that block by $kck$ shortens it by four. Iterating, every word equals an alternating word of length at most $7$. There are exactly two alternating words of each length from $1$ to $7$ and one of length $0$, and the length-$7$ word beginning with $k$ is $kckckck = kck$ by claim 1; the remaining fourteen are those listed in the statement.

In $\mathbb{R}$ with $A = (0,1) \cup (1,2) \cup \{3\} \cup ([4,5] \cap \mathbb{Q})$: by [A2] the closure of the four-term union is the union of the four closures, which by steps 2.1, 3.1 and 3.2 are $[0,1]$, $[1,2]$, $\{3\}$ and $[4,5]$; hence $kA = [0,2] \cup \{3\} \cup [4,5]$.

$cA = (-\infty,0] \cup \{1\} \cup [2,3) \cup (3,4) \cup ([4,5] \setminus \mathbb{Q}) \cup (5,\infty)$. Here $\overline{[2,3)} = [2,3]$, because $[2,3]$ is closed and contains $[2,3)$ while monotonicity gives $[2,3] = \overline{(2,3)} \subseteq \overline{[2,3)}$; the other five closures are $(-\infty,0]$, $\{1\}$, $[3,4]$, $[4,5]$ and $[5,\infty)$ by steps 2.1, 3.1 and 3.2. So by [A2] the closure of the six-term union is $(-\infty,0] \cup \{1\} \cup [2,3] \cup [3,4] \cup [4,5] \cup [5,\infty) = (-\infty,0] \cup \{1\} \cup [2,\infty)$, that is $kcA = (-\infty,0] \cup \{1\} \cup [2,\infty)$.

$ckA = (-\infty,0) \cup (2,3) \cup (3,4) \cup (5,\infty)$, and its closure is $(-\infty,0] \cup [2,4] \cup [5,\infty)$ by [A2] and steps 3.2 and 2.1; so $kckA = (-\infty,0] \cup [2,4] \cup [5,\infty)$.

$ckcA = (0,1) \cup (1,2)$, whose closure is $[0,2]$ by [A2] and step 3.2; so $kckcA = [0,2]$. Then $ckckcA = (-\infty,0) \cup (2,\infty)$, whose closure is $(-\infty,0] \cup [2,\infty)$ by [A2] and step 3.2; so $kckckcA = (-\infty,0] \cup [2,\infty)$.

$ckckA = (0,2) \cup (4,5)$, whose closure is $[0,2] \cup [4,5]$ by [A2] and step 3.2; so $kckckA = [0,2] \cup [4,5]$.

The seven sets $A$, $kA$, $kckA$, $kckckA$, $kcA$, $kckcA$, $kckckcA$ have the following membership pattern at the five test points $0$, $1$, $3$, $9/2$, $6$, writing $1$ for "belongs" and $0$ for "does not": $A$ gives $(0,0,1,1,0)$, since $9/2$ is a rational in $[4,5]$; $kA$ gives $(1,1,1,1,0)$; $kckA$ gives $(1,0,1,0,1)$; $kckckA$ gives $(1,1,0,1,0)$; $kcA$ gives $(1,1,1,1,1)$; $kckcA$ gives $(1,1,0,0,0)$; $kckckcA$ gives $(1,0,1,1,1)$.

The remaining seven words of claim 2 are the complements of these seven — the list of fourteen words consists of the seven above and those seven preceded by $c$ — so their patterns are the bitwise complements $(1,1,0,0,1)$, $(0,0,0,0,1)$, $(0,1,0,1,0)$, $(0,0,1,0,1)$, $(0,0,0,0,0)$, $(0,0,1,1,1)$, $(0,1,0,0,0)$. The fourteen patterns are pairwise distinct, so the fourteen sets are, and claim 3 holds.