Showing this function on the Cantor set is onto [0,1] The excerpt below is taken from Rosenthal's A First Look at Rigorous Probability. $K$ refers to the cantor set.

My question refers to the statement "It is easily checked that $f(K) =[0,1]$. I am thinking that this can by proved by taking any number in $[0,1]$, writing the binary expansion for it (that is, write it of the form $\sum_{n=1}^\infty b_n\cdot2^{-n},\ b_n\in \{0,1\}$ and then show that there is a point in the cantor set that will give $d_n = b_n \forall n$. How would I do this last step? That is, how would I show that such a point exists in the cantor set, $K$?
To state the question again: how I would show that there is a $y\in K$ which corresponds to some point in $[0,1]$?
A secondary question is: Can we show that $f(K) = [0,1]$, where $f$ is as in the attached image, without explicitly using binary/ternary expansions, and preferably also not using compactness? It is not important that this secondary question is answered: if the first question is answered and this one is not, I will select an answer for the question and then probably just create a separate question eventually for this and add a bounty if necessary.
Thank you.
Note: This question is technically addressed here, but the answers seems to say to me "here is a function, it is a surjection, without explaining why it is a surjection.
 A: The quoted argument is probably intended to be filled in as follows. Let $x\in[0,1]$; then $x$ has a binary expansion
$$x=\sum_{k\ge 1}\frac{b_k}{2^k}\;,$$
where each $b_k\in\{0,1\}$. The sequence $\langle b_k:k\in\Bbb Z^+\rangle$ now defines a nest of closed intervals as follows.
Let $I_0=[0,1]$. Given $I_k$ for some $k\in\Bbb N$, let $I_{k+1}$ be the closed left third of $I_k$ if $b_{k+1}=0$, and let $I_{k+1}$ be the right closed third of $I_k$ if $b_{k+1}=1$. The sequence $\langle I_k:k\in\Bbb N\rangle$ is a decreasing nest of closed intervals, so $\bigcap_{k\in\Bbb N}I_k\ne\varnothing$. On the other hand, the length of $I_k$ is $3^{-k}$, so the diameter of $\bigcap_{k\in\Bbb N}I_k$ is $0$, and it follows that there is a unique $y\in\bigcap_{k\in\Bbb N}I_k$. 
All that remains is to verify that $x=f(y)$, which can be done by showing that $b_k=d_k(y)$ for each $k\in\Bbb Z^+$. But this is clear from the construction of the intervals $I_k$: for each $k\in\Bbb N$ we have $d_{k+1}(y)=0$ iff $y$ is to the left of the nearest open interval removed at step $k+1$ iff $I_{k+1}$ is the closed left third of $I_k$ iff $b_{k+1}=0$.
