The lines given by $$f_1:\mathbb{W}\rightarrow \mathbb{N},~~f_1(n)=(6n+1)\cdot2^{k}$$ $$f_2:\mathbb{W}\rightarrow \mathbb{N},~~f_2(n)=(6n+3)\cdot2^{k}$$ $$f_3:\mathbb{W}\rightarrow \mathbb{N},~~f_3(n)=(6n+5)\cdot2^{k}$$ where $n,k$ are arbitrary whole numbers.
By ignoring the $2^{k}$ I can see that any given odd number will appear on one of the lines, and when multiplied by a factor of 2 will give any even number. How do you suppose I would go about and prove all of the natural numbers exist on the lines?
Your observation is pretty much correct, you just need to formalise it. You can start by proving that every odd $x$ is in one of the sequences, probably by assuming that there's an $x$ not in $f_1$ or $f_2$, and hence showing that it must be in $f_3$. After that, if you can then prove that any even number can be written as $2^k\cdot x$ for some odd $x$, you're done.