Cantor construction is continuous

Denote the funcion you constructed on $\mathbb{R}\setminus C$ by $f$. The question is how to define Cantor's function on $[0,1]$. You can use the formula $$ F(x)=\sup_{t\in[0,x]\setminus C} f(t),\qquad x\in[0,1] $$ The function $F$ is continuous thanks to the following lemma.

Lemma. Let $F:[a,b]\to\mathbb{R}$ is non decreasing and $F([a,b])$ is dense in $[F(a),F(b)]$, then $F$ is continuous.

Proof. Since $F$ is non decreasing there exist finite one sided limits at any $x\in[a,b]$, denote them $F(x+)$ and $F(x-)$. Assume $F$ is discontinuous at $c\in[a,b]$, then $f(c-)<f(c+)$. In this case $F([a,b])\cap[F(c-),F(c+)]=\{F(c)\}$, so $F([a,b])$ is not dense in $[F(a),F(b)]$. Contradiction.