The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set

Comparison of two digit sequences. Let $a \ne b$ in $D$ and let $k$ be the least index with $a_k \ne b_k$, which exists by [L5]; suppose $a_k = 0$ and $b_k = 2$. Then $\Phi(b) - \Phi(a) = \sum_{j \ge 0}(b_j - a_j)3^{-j-1}$ by [L2], the terms with $j < k$ vanish, and the tail $R := \sum_{j \ge k+1}(b_j - a_j)3^{-j-1}$ satisfies $|R| \le \sum_{j \ge k+1} 2 \cdot 3^{-j-1} = 3^{-k-1}$ by [L2], since $|b_j - a_j| \le 2$; hence $\Phi(b) - \Phi(a) \ge 2 \cdot 3^{-k-1} - 3^{-k-1} = 3^{-k-1} > 0$. The same computation with the halved digits gives $\gamma(\Phi(b)) - \gamma(\Phi(a)) = 2^{-k-1} + R'$ with $|R'| \le \sum_{j \ge k+1} 2^{-j-1} = 2^{-k-1}$, so $\gamma(\Phi(b)) \ge \gamma(\Phi(a))$. Consequently, for $s, t \in C$ with $s \le t$ one has $\gamma(s) \le \gamma(t)$: this is trivial if $s = t$, and otherwise the least index $k$ at which the digit sequences differ must have the digit of $t$ equal to $2$, by the first computation applied both ways.

Values at the endpoints. The constant sequence $\bar 0$ has $\Phi(\bar 0) = 0$ and $\gamma(0) = 0$; the constant sequence $\bar 2$ has $\Phi(\bar 2) = \sum_{k \ge 0} 2 \cdot 3^{-k-1} = 1$ and $\gamma(1) = \sum_{k \ge 0} 2^{-k-1} = 1$, by [L2]. Both $0$ and $1$ lie in $C$ by [L3].

Claims 1 and 2. For $x \in [0,1]$ the set $A_x := \{\gamma(t) : t \in C,\ t \le x\}$ is nonempty and bounded above by $1$ by [L1], so $c(x) = \sup A_x$ exists, is unique and lies in $[0,1]$ by [L1] and [L4]; that is claim 1 apart from the extension property. If $0 \le x \le y \le 1$ then $A_x \subseteq A_y$, so $c(x) \le c(y)$ by [L4], which is claim 2. And for $t \in C$: $\gamma(t) \in A_t$, while $\gamma(t)$ is an upper bound of $A_t$ by step 1.1, so $\gamma(t) = \sup A_t = c(t)$ by [L4].

The two endpoints of a gap carry the same value of $\gamma$. Let $u < v$ with $u, v \in C$ and $(u,v) \cap C = \varnothing$, and put $a := \Phi^{-1}(u)$, $b := \Phi^{-1}(v)$, with $k$ the least index where they differ; by step 1.1 and $u < v$ we have $a_k = 0$ and $b_k = 2$. If some $j > k$ had $a_j = 0$, let $a'$ agree with $a$ except that $a'_j = 2$; then $\Phi(a') \in C$, $\Phi(a') > u$ by step 1.1, and $a'$ still differs from $b$ first at $k$ with $a'_k = 0 < 2 = b_k$, so $\Phi(a') < v$ by step 1.1, putting $\Phi(a')$ in $(u,v) \cap C$, which is empty. Hence $a_j = 2$ for every $j > k$. Symmetrically, if some $j > k$ had $b_j = 2$, replacing it by $0$ gives $b'$ with $\Phi(b') < v$ and $\Phi(b') > u$, again impossible; hence $b_j = 0$ for every $j > k$. Writing $P := \sum_{j<k}(a_j 2^{-1})2^{-j-1} = \sum_{j<k}(b_j 2^{-1})2^{-j-1}$, [L2] now gives $\gamma(u) = P + 0 + \sum_{j \ge k+1} 2^{-j-1} = P + 2^{-k-1}$ and $\gamma(v) = P + 2^{-k-1} + 0 = P + 2^{-k-1}$, so $\gamma(u) = \gamma(v)$.

Claim 4, first half. Let $u < v$ with $u,v \in C$ and $(u,v) \cap C = \varnothing$, and let $x \in [u,v]$. Every $t \in C$ with $t \le x$ satisfies $t \le u$ or $t = v$: indeed if $t > u$ then $t \le x \le v$ and $t \notin (u,v)$ force $t = v$. In the first case $\gamma(t) \le \gamma(u)$ by step 1.1, and in the second $\gamma(t) = \gamma(v) = \gamma(u)$ by step 2.2. So $\gamma(u)$ is an upper bound of $A_x$ and belongs to it, whence $c(x) = \gamma(u)$ by [L4]: $c$ is constant on $[u,v]$, with the value $c(u)$ given by step 2.1.

Claim 3. Let $s \in [0,1]$. Let $T : \mathbb{R} \to \mathbb{R}$ be $T(r) := 2r$ for $r < 2^{-1}$ and $T(r) := 2r - 1$ for $r \ge 2^{-1}$, a definition by cases on the total order, and by [L5] let $(r_n)$ satisfy $r_0 = s$ and $r_{n+1} = T(r_n)$; put $\beta_n := 0$ when $r_n < 2^{-1}$ and $\beta_n := 1$ otherwise, so $r_{n+1} = 2r_n - \beta_n$. An induction ([L5]) gives $r_n \in [0,1]$ for every $n$, since $0 \le r < 2^{-1}$ gives $0 \le 2r < 1$ and $2^{-1} \le r \le 1$ gives $0 \le 2r - 1 \le 1$ by [L9]; a second induction gives $s = \sum_{k<n}\beta_k 2^{-k-1} + 2^{-n} r_n$ for every $n$, the step being $\sum_{k<n+1}\beta_k2^{-k-1} + 2^{-n-1}r_{n+1} = \sum_{k<n}\beta_k2^{-k-1} + \beta_n 2^{-n-1} + 2^{-n-1}(2r_n - \beta_n) = \sum_{k<n}\beta_k2^{-k-1} + 2^{-n}r_n$. Hence $0 \le s - \sum_{k<n}\beta_k2^{-k-1} \le 2^{-n}$, so by [L6] the partial sums converge to $s$ and $s = \sum_{k \ge 0}\beta_k 2^{-k-1}$. Now $a := (2\beta_k)_k$ lies in $D$, the point $x := \Phi(a)$ lies in $C$ by [L1], and $\gamma(x) = \sum_k \beta_k 2^{-k-1} = s$; by step 2.1, $c(x) = \gamma(x) = s$. With step 1.2 and step 2.1 this also gives $c(0) = \gamma(0) = 0$ and $c(1) = \gamma(1) = 1$.

Claim 4, second half. Let $x \in [0,1] \setminus C$. The set $A := \{t \in C : t \le x\}$ is nonempty by [L3] and bounded above by $x$, so $u := \sup A$ exists by [L4]; by [L4] every $N_\varepsilon(u)$ meets $A \subseteq C$, so $u \in \overline{C} = C$ by [L3], and $u \le x$ with $u \ne x$, so $u < x$. The set $B := \{t \in C : t \ge x\}$ is nonempty by [L3], since $1 \in C$ and $x \le 1$, and is bounded below by $x$, so $v := \inf B$ exists by [L4]; likewise $v \in C$ and $v > x$. If $t \in C$ satisfied $u < t < v$, then $t \le x$ would put $t \in A$ and force $t \le u$, while $t \ge x$ would put $t \in B$ and force $t \ge v$, and one of the two holds by totality of the order ([L9]); so $(u,v) \cap C = \varnothing$. By step 3.1 the function $c$ is constant on $[u,v]$, and $N_\delta(x) \subseteq (u,v)$ for $\delta := \min\{x - u,\ v - x\} > 0$ by [L7], [L8] and [L9].

Claims 1 and 2 are step 2.1, claim 3 is step 3.2, and claim 4 is steps 3.1 and 4.1 together; so all four hold.