Well, as Gerry has pointed out, this is certainly not true for all $\alpha$. On the other hand, this is true for a.e. $\alpha$. More precisely, the sequence $2^n\alpha$ is equidistributed mod 1 for a.e. $\alpha$.

I believe this result is due to H. Weyl and can be found in Cornfeld, Fomin and Sinai `Ergodic Theory'. (I don't have it with me.) 

The same must be true for the Fibonacci sequence, I'm sure. 

So, what you probably need is for this to be true for all $\alpha$, except some small (countable?) set. After all, $\|2^n\alpha\|<\varepsilon$ is indeed much weaker than equidistibution.  

**Update.** Come to think about it, the answer is as follows: let 
$$
\alpha = \sum_{k=1}^\infty a_k2^{-k}
$$
be the binary expansion of $\alpha$. Then the sequence $2^n\alpha\bmod 1$ gets arbitrarily close to 0 if and only if the sequence $(a_k)$ has unbounded strings of 0s. In particular, any rational $\alpha$ is out of the picture, apart from the binary rationals, of course.

All in all, your set of $\alpha$'s is indeed of full measure, but the exceptional set is of Hausdorff dimension 1, i.e., pretty big.

For the Fibonacci sequence you'll need to replace binary expansion with the $\beta$-expansion, where $\beta=(\sqrt5-1/)2$, with the same conclusion.