You don’t have to show that $f$ is continuous in order to conclude that $\mathcal{C}$ is uncountable: that follows from the fact that $f$ is surjective, assuming that you know that $[0,1]$ is uncountable. Your argument for surjectivity is correct, though it could be stated a bit better, but for clarity you ought to deal with a point raised by Dan Brumleve. Here’s your argument, slightly restated:
Let $y\in [0,1]$ be arbitrary, and let $(0.y_1y_2\dots)$ be a binary representation of $y$. For each positive integer $i$ let $a_i=2y_i$, and let $x$ be the number whose ternary representation is $(0.a_1a_2\dots)$; then $y=f(x)$, so $f$ is surjective.
Since this looks at first sight as if you were actually constructing a function from $[0,1]$ to $\mathcal{C}$, it would help the reader if you were to point out that this isn’t the case. For example, $y=1/2$ has two binary representations, $0.1\bar{0}$ and $0.0\bar{1}$, where the bar indicates a repeating digit, so it’s both $f(0.2_{\text{three}})=$ $f(\frac23)$ and $f(0.\bar{2})=f(\frac13)$.
As I said, continuity of $f$ isn’t needed if all you want is to prove the uncountability of $\mathcal{C}$, but if for some other reason you have to prove the continuity of $f$, here’s one approach. For $x\in\mathbb{R}$ and $r>0$ let $B(x,r)=\{y\in\mathbb{R}:\vert y-x\vert\le r\}$. Suppose that $0.x_1x_2\dots$ is the $1$-less ternary expansion of some $x\in\mathcal{C}$. For $n\in\mathbb{Z}^+$ let $T_n(x)=$ $\{(0.a_1a_2\dots)\in\mathcal{C}:\forall i\le n [a_i=x_i]\}$. Show that $$B(x,3^{-(n+1)})\cap\mathcal{C}\subseteq T_n(x)\subseteq B(x,3^{-n})\cap\mathcal{C}.$$ Then show that for each $r>0$ there is an $n(r)\in\mathbb{Z}^+$ such that $T_{n(r)}(x)\subseteq B(f(x),r)$.
Another approach is to show that $f$ preserves convergent sequences, i.e., that if $\langle x_n:n\in\mathbb{N}\rangle$ is a sequence in $\mathcal{C}$ that converges to $x\in\mathcal{C}$, then $\langle f(x_n):n\in\mathbb{N}\rangle \to f(x)$ in $[0,1]$. A good first step would be to show that if $\langle x_n:n\in\mathbb{N}\rangle$ is a sequence in $\mathcal{C}$ that converges to $x\in\mathcal{C}$, then for each $m\in\mathbb{Z}^+$ there is an $n(m)\in\mathbb{N}$ such that $x_n\in T_m(x)$ whenever $n\ge n(m)$: that is, every term of the sequence from $x_{n(m)}$ on agrees with $x$ to at least $m$ ternary places. Then each term of $\langle f(x_n):n\in\mathbb{N}\rangle$ from $f(x_{n(m)})$ on agrees with $f(x)$ to at least $m$ binary places. From this it’s not hard to conclude that $\langle f(x_n):n\in\mathbb{N}\rangle \to f(x)$.
All you need is $EY_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$ as $n \to \infty$ and this follows from DCT. [$Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} $ is dominated by $Y_1^{2}$ which is integrable. Of course, the events $\{|Y_1| >\epsilon \sigma \sqrt n\}$ decrease to empty set so $Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$ almost surely. ].
Best Answer
Surjectivity of $f$ is equivalent to proof that every $y\in [0,1]$ has a base-$2$ representation. We can obtain such a representation inductively by constructing boxed intervals of lenght $2^{-n}$ converging to $y$.
Explicitly we start from $[0,1]$, divide it in half and take the one cointaining $y$, formally it's better to split $[0,1]$ in $[0,1/2)$ and $[1/2,1]$ so they don't intersect. Next we repeat the construction and get a sequence of intervals. Each choice "left-right" is equivalent a choice "0-1", so we obtain an element $x$ of $X$, we need to prove that $f(x)=y$. Notice that if we stop the sum defining $f(x)$ at time $n$ we obtain the left extrem of the $n$-th interval constructed above, I leave you the details.
The map is not open, take a look at some open ball around $x=(1,0,0\dots)$ of radius less than $1/2$.
Hope it is clear, if not, please ask.
Edit
I will make more explicit the argument to show that $f$ is not open. As said above consider an open ball $B$ around $x=(1,0,0\dots)$ of small radius, say $1/2$, and take any $y\in B$. Being $d(x,y)<1/2$ we must have $y=(1,y_2, \dots)$, this implies that $f(y)\geq 1/2=f(x)$ and hence $f(B)$ must be contained in $[1/2,1]$ which is not open with the euclidean topology.