The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points

$C$ is compact, claim 1. First, for $\lambda \ne 0$ and $c \in \mathbb{R}$ the set $\lambda S + c := \{\lambda s + c : s \in S\}$ is closed whenever $S$ is: if $x \notin \lambda S + c$ then $(x - c)\lambda^{-1} \notin S$, so by [L3] there is a real $\eta > 0$ with $N_\eta((x-c)\lambda^{-1}) \cap S = \varnothing$, and every $z$ with $|z - x| < |\lambda|\eta$ satisfies $|(z-c)\lambda^{-1} - (x-c)\lambda^{-1}| = |z-x| \cdot |\lambda|^{-1} < \eta$ by [L10] and [L12], hence $(z-c)\lambda^{-1} \notin S$ and $z \notin \lambda S + c$. Now every $C_n$ is closed, by induction on $n$ ([L11]): $C_0 = [0,1]$ is closed by [L3], and $C_{n+1}$ is the union of the two closed sets $\tfrac13 C_n$ and $\tfrac23 + \tfrac13 C_n$, hence closed by [L4]. So $C = \bigcap_n C_n$ is closed by [L4], and $C \subseteq [0,1]$ is bounded by [L1] and [L3]; by [L5] it is compact.

$C$ has content zero and measure zero, claim 2. By induction on $n$ ([L11]) the following holds for every $n$: there are $m \in \mathbb{N}$ and reals $u_0 \le v_0, \dots, u_m \le v_m$ with $C_n \subseteq \bigcup_{j \le m}[u_j,v_j]$ and $\sum_{j \le m}(v_j - u_j) = (2 \cdot 3^{-1})^{n}$. At $n = 0$ take the single interval $[0,1]$, of total length $1 = (2 \cdot 3^{-1})^0$ by [L1]. Given such a list at $n$, define $2m + 2$ intervals by $[u_j 3^{-1},\, v_j 3^{-1}]$ for $j \le m$ and $[2 \cdot 3^{-1} + u_{j-m-1}3^{-1},\, 2 \cdot 3^{-1} + v_{j-m-1}3^{-1}]$ for $m < j \le 2m+1$; they cover $\tfrac13 C_n$ and $\tfrac23 + \tfrac13 C_n$ respectively, hence cover $C_{n+1}$, and their total length is $3^{-1}(2 \cdot 3^{-1})^{n} + 3^{-1}(2 \cdot 3^{-1})^{n} = (2 \cdot 3^{-1})^{n+1}$ by [L11] and [L12]. Since $0 < 2 \cdot 3^{-1} < 1$ by [L12], [L10] gives, for every real $\varepsilon > 0$, an $n$ with $(2 \cdot 3^{-1})^{n} \le \varepsilon$; as $C \subseteq C_n$ by [L1], the corresponding finite list covers $C$ with total length at most $\varepsilon$. So $C$ has content zero by [L6], and hence measure zero by [L6].

$C$ is perfect, claim 3. $C$ is closed by step 1.1. Let $x \in C$ and let the real $\varepsilon > 0$ be given. By [L2] write $x = \Phi(a)$ with $a \in D$. By [L10] and [L12] fix $k \in \mathbb{N}$ with $2 \cdot 3^{-k-1} < \varepsilon$, and define $b \in D$ by $b_j := a_j$ for $j \ne k$ and $b_k := 2 - a_k$, so $b_k \in \{0,2\}$ and $b \ne a$. Then $\Phi(b) \in C$ and $\Phi(b) \ne \Phi(a)$ by [L2], while $\Phi(b) - \Phi(a) = \sum_{j \ge 0}(b_j - a_j)3^{-j-1} = (b_k - a_k)3^{-k-1}$ by [L2], all other terms being $0$, so $|\Phi(b) - x| = 2 \cdot 3^{-k-1} < \varepsilon$ by [L10]. Thus $N_\varepsilon(x)$ contains a point of $C$ other than $x$, for every $\varepsilon$, so $x$ is not isolated in $C$; by [L7] $C$ is perfect.

$C$ contains no nondegenerate interval and is nowhere dense, claim 5. By step 1.2 the set $C$ is null, so by [L6] it contains no $[u,v]$ with $u < v$; in particular it contains no interval of any of the four bounded forms with distinct endpoints, since such an interval contains a closed one with distinct endpoints by [L6] and [L12]. Its interior is therefore empty: if $N_\varepsilon(x) \subseteq C$ for some real $\varepsilon > 0$, then $[x - \varepsilon \cdot 2^{-1},\, x + \varepsilon \cdot 2^{-1}] \subseteq N_\varepsilon(x) \subseteq C$ by [L3] and [L12], an interval with distinct endpoints. Since $C$ is closed by step 1.1, it equals its closure, so [L8] gives that $C$ is nowhere dense.

$C$ is uncountable, claim 4. $C$ is nonempty, since $0 \in C$ by [L1], and perfect by step 2.1, so [L7] applies.

Connected subsets, claim 6. Let $E \subseteq C$ be connected and nonempty. By [L9] $E$ is order-convex, so if $u, v \in E$ with $u < v$ then $[u,v] \subseteq E \subseteq C$, contradicting step 2.2. Hence no two distinct elements of $E$ exist, and $E$, being nonempty, is a single point.

Claims 1 to 6 are steps 1.1, 1.2, 2.1, 3.1, 2.2 and 3.2 respectively, so all six hold.